Okay, here we have this:
Considering the provided information we are going to calculate what is the probability that the two candies selected are taffy. So, for this, first we are going to calculate the probability that the first is taffy, and then the probability that the second is taffy. Finally we will multiply these two probabilities to find the total probability.
Remember that the simple probability of an event is equal to favorable events, over possible events.
First is taffy:
At the beginning there are 14 sweets, and 3 are taffy, so there are 3 favorable events and 14 possible, then:
First is taffy=3/14
Second is taffy:
Now, in the bag there are 13 sweets left, and of those 2 are taffy, so now there are 2 favorable events out of 13 possible:
Second is taffy=2/13
The first and second are taffy:
First is taffy*Second is taffy=3/14*2/13
First is taffy*Second is taffy=3/91
First is taffy*Second is taffy=0.033
First is taffy*Second is taffy=3.3%
Finally we obtain that the probability that the two candies selected are taffy is aproximately 0.033 or 3.3%.
Suppose the coordinate of p=2 and PQ=8. Whare are the possible midpoints for PQ?
The midpoint for segment PQ can be calculated as:
[tex]\frac{P+Q}{2}[/tex]Then, the midpoint of PQ is:
[tex]\frac{2\text{ + Q}}{2}=1+0.5Q[/tex]Additionally, PQ can be calculated as:
[tex]PQ=\left|Q-P\right|[/tex]So:
[tex]\begin{gathered} \left|Q-P\right|=8 \\ \left|Q-2\right|=8 \end{gathered}[/tex]It means that:
[tex]\begin{gathered} Q-2=8\text{ or } \\ 2\text{ - Q = 8} \end{gathered}[/tex]Solving for Q, we get:
Q = 8 + 2 = 10 or Q = 2 - 8 = -6
Finally, replacing these values on the initial equation for the midpoint, we get:
If Q = 10, then:
midpoint = 1 + 0.5(10) = 1 + 5 = 6
If Q = -6, then:
midpoint = 1 + 0.5(-6) = 1 - 3 = -2
The possible midpoints for PQ are 6 and -2
Find the exact solution to the exponential equation. (No decimal approximation)
Let's solve the equation:
[tex]\begin{gathered} 54e^{3x+3}=16 \\ e^{3x+3}=\frac{16}{54} \\ e^{3x+3}=\frac{8}{27} \\ \ln e^{3x+3}=\ln (\frac{8}{27}) \\ 3x+3=\ln (\frac{2^3}{3^3}) \\ 3x+3=\ln (\frac{2}{3})^3 \\ 3x+3=3\ln (\frac{2}{3}) \\ 3x=-3+3\ln (\frac{2}{3}) \\ x=-1+\ln (\frac{2}{3}) \\ x=-1+\ln 2-\ln 3 \end{gathered}[/tex]Therefore the solution of the equation is:
[tex]x=-1+\ln 2-\ln 3[/tex]Please help me no other tutor could or understand it
We must find the equation that models the amount of medication in the bloodstream as a function of the days passed from the initial dose. The initial dose is a and we are going to use x for the number of days and M for the amount of mediaction in the bloodstream. We are going to model this using an exponential function which means that the variable x must be in the exponent of a power:
[tex]M(x)=a\cdot b^x[/tex]We are told that the half-life of the medication is 6 hours. This means that after 6 hours the amount of medication in the bloodstream is reduced to a half. If the initial dose was a then the amount after 6 hours has to be a/2. We are going to use this to find the parameter b but first we must convert 6 hours into days since our equation works with days.
Remember that a day is composed of 24 hours so 6 hours is equivalent to 6/24=1/4 day. This means that the amount of medication after 1/4 days is the half of the initial dose. In mathematical terms this means M(1/4)=M(0)/2:
[tex]\begin{gathered} \frac{M(0)}{2}=M(\frac{1}{4}) \\ \frac{a\cdot b^0}{2}=a\cdot b^{\frac{1}{4}} \\ \frac{a}{2}=a\cdot b^{\frac{1}{4}} \end{gathered}[/tex]We can divide both sides of this equation by a:
[tex]\begin{gathered} \frac{\frac{a}{2}}{a}=\frac{a\cdot b^{\frac{1}{4}}}{a} \\ \frac{1}{2}=b^{\frac{1}{4}} \end{gathered}[/tex]Now let's raised both sides of this equation to 4:
[tex]\begin{gathered} (\frac{1}{2})^4=(b^{\frac{1}{4}})^4 \\ \frac{1}{2^4}=b^{\frac{1}{4}\cdot4} \\ b=\frac{1}{16} \end{gathered}[/tex]Which can also be written as:
[tex]b=16^{-1}[/tex]Then the equation that models how much medication will be in the bloodstream after x days is:
[tex]M(x)=a\cdot16^{-x}[/tex]Using this we must find how much medication will be in the bloodstream after 4 days for an initial dose of 500mg. This basically means that a=500mg, x=4 and we have to find M(4):
[tex]M(4)=500mg\cdot16^{-4}=0.00763mg[/tex]So after 4 days there are 0.00763 mg of medication in the bloodstream.
Now we have to indicate how much more medication will be if the initial dose is 750mg instead of 500mg. So we take a=750mg and x=4:
[tex]M(4)=750mg\cdot16^{-4}=0.01144mg[/tex]If we substract the first value we found from this one we obtained the required difference:
[tex]0.01144mg-0.00763mg=0.00381mg[/tex]So the answer to the third question is 0.00381mg.
-27\sqrt(3)+3\sqrt(27), reduce the expression
Explanation
[tex]-27\sqrt[]{3}+3\sqrt[]{27}[/tex]Step 1
Let's remember one propertie of the roots
[tex]\sqrt[]{a\cdot b}=\sqrt[]{a}\cdot\sqrt[]{b}[/tex]hence
[tex]\sqrt[]{27}=\sqrt[]{9\cdot3}=\sqrt[]{9}\cdot\sqrt[]{3}=3\sqrt[]{3}[/tex]replacing in the expression
[tex]\begin{gathered} -27\sqrt[]{3}+3\sqrt[]{27} \\ -27\sqrt[]{3}+3(3\sqrt[]{3}) \\ -27\sqrt[]{3}+9\sqrt[]{3} \\ (-27+9)\sqrt[]{3} \\ -18\sqrt[]{3} \end{gathered}[/tex]therefore, the answer is
[tex]-18\sqrt[]{3}[/tex]I hope this helps you
=Given f(x) = -0.4x – 10, what is f(-12)? If it does not exist,enter DNE.
We have the function:
[tex]f\mleft(x\mright)=-0.4x-10[/tex]And we need to find its value when x = -12. So, replacing x with -12, we obtain:
[tex]f(-12)=-0.4(-12)-10=4.8-10=-5.2[/tex]Notice that the product of two negative numbers is a positive number.
Therefore, the answer is -5.2.
Solve this equation 3n+8=20
Given the equation below
[tex]3n\text{ + 8 = 20}[/tex]Step 1
Collect like terms.
[tex]\begin{gathered} 3n=20-8 \\ 3n=12 \end{gathered}[/tex]Step 2
Divide both sides of the equation obtained, by the coefficient of the unknown.
[tex]\begin{gathered} \text{The unknown is n.} \\ \text{The co}efficient\text{ of n is 3.} \\ \text{Thus,} \\ \frac{3n}{3}=\frac{12}{3} \\ \Rightarrow n=4 \end{gathered}[/tex]Hence, the value of n in the equation is 4
3. f(x) = |-3x - 1|3. For this function, findeach of the following:a. f(-1)b. f(0)c. f(3)
Given the absolute function;
[tex]f(x)=|-3x-1|[/tex](a)
[tex]\begin{gathered} f(-1)=|-3x-1| \\ f(-1)=|-3(-1)-1| \\ f(-1)=|3-1| \\ f(-1)=|2| \\ f(-1)=2 \end{gathered}[/tex](b)
[tex]\begin{gathered} f(0)=|-3(0)-1| \\ f(0)=|0-1| \\ f(0)=|-1| \end{gathered}[/tex]Here, we recall the absolute rule that;
[tex]|-a|=a[/tex]Thus, we have;
[tex]f(0)=|-1|=1[/tex](c)
[tex]\begin{gathered} f(3)=|-3(3)-1| \\ f(3)=|-9-1| \\ f(3)=|-10| \\ f(3)=10 \end{gathered}[/tex]Rewrite the fraction with a rational denominator:
[tex]\frac{1}{\sqrt{5} +\sqrt{3} -1}[/tex]
Give me a clear and concise explanation (Step by step)
I will report you if you don't explain
The expression with rational denominator is [tex]\frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]
How to rewrite the fraction?From the question, the fraction is given as
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1}[/tex]
To rewrite the fraction with a rational denominator, we simply rationalize the fraction
When the fraction is rationalized, we have the following equation
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{1}{\sqrt 5 + \sqrt{3} - 1} \times \frac{\sqrt 5 - \sqrt{3} + 1}{\sqrt 5 - \sqrt{3} + 1}[/tex]
Evaluate the products in the above equation
So, we have
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{(\sqrt 5)^2 - (\sqrt{3} + 1)^2}[/tex]
This gives
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{5 - 10 - 2\sqrt 3}[/tex]
So, we have
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{- 5 - 2\sqrt 3}[/tex]
Rationalize again
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{- 5 - 2\sqrt 3} \times \frac{- 5+2\sqrt 3}{- 5 +2\sqrt 3}[/tex]
This gives
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{(-5)^2 - (2\sqrt 3)^2}[/tex]
So, we have
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{25 -12}[/tex]
Evaluate
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]
Hence, the expression is [tex]\frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]
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ExpenseYearly costor rateGasInsuranceWhat is the cost permile over the course ofa year for a $20,000 carthat depreciates 20%,with costs shown in thetable, and that hasbeen driven for 10,000miles?$425.00$400.00$110,00$100.0020%OilRegistrationDepreciationB. $4.10 per mileA. $1.10 per mileC. $0.25 per mileD. $0.50 per mile
Step 1:
Find the depreciation
[tex]\begin{gathered} \text{Depreciation = 20\% of \$20,000} \\ \text{Depreciation = }\frac{20}{100}\text{ }\times\text{ \$20000} \\ \text{Depreciation = \$4000} \end{gathered}[/tex]Step 2:
Total cost = $425 + $400 + $110 + $100 + $4000
Total cost = $5035
Final answer
[tex]\begin{gathered} \text{Cost per mile = }\frac{5035}{10000} \\ \text{Cost per mile = \$0.5035} \\ \text{Cost per mile = \$0.50} \end{gathered}[/tex]Option D $0.50 per mile
The cost, c(x) in dollars per hour of running a trolley at an amusement park is modelled by the function [tex]c(x) = 2.1x {}^{2} - 12.7x + 167.4[/tex]Where x is the speed in kilometres per hour. At what approximate speed should the trolley travel to achieve minimum cost? A. About 2km/h B about 3km/h C about 4km/D about 5km/hr
The equation is modelled by the function,
c(x) = 2.1x^2 - 12.7x + 167.4
The general form of a quadratic equation is expressed as
ax^2 + bx + c
The given function is quadratic and the graph would be a parabola which opens upwards because the value of a is positive
Since x represents the speed, the speed at which the he
Find the sum of the arithmetic series -1+ 2+5+8+... where n=7.A. 56B. 184C. 92D. 380Reset Selection
The arithmetic series is:
-1 + 2 + 5 + 8 + .....
The first term, a = -1
The common difference, d = 2 - (-1)
d = 3
The number of terms, n = 7
Find the sum of the arithmetic series below
[tex]\begin{gathered} S_n=\frac{n}{2}[2a+(n-1)d] \\ \\ S_7=\frac{7}{2}[2(-1)+(7-1)(3)] \\ \\ S_7=\frac{7}{2}(-2+18) \\ \\ S_7=\frac{7}{2}(16) \\ \\ S_7=56 \end{gathered}[/tex]Therefore, the sum of the arithmetic series = 56
I need help to find the indicated operation:g(x)= -x^2 +4xh(x)= -4x-1Find (3g-h)(-3)
We have the following functions:
[tex]\begin{gathered} g\mleft(x\mright)=-x^2+4x \\ h\mleft(x\mright)=-4x-1 \end{gathered}[/tex]And we need to find:
[tex](3g-h)(-3)[/tex]Step 1. Find 3g by multiplying g(x) by 3:
[tex]\begin{gathered} g(x)=-x^2+4x \\ 3g=3(-x^2+4x) \end{gathered}[/tex]Use the distributive property to multiply 3 by the two terms inside the parentheses:
[tex]3g=-3x^2+12x[/tex]Step 2. Once we have 3g, we subtract h(x) to it:
[tex]3g-h=-3x^2+12x-(-4x-1)[/tex]Here we have 3g and to that, we are subtracting h which in parentheses.
Simplifying the expression by again using the distributive property and multiply the - sign by the two terms inside the parentheses:
[tex]3g-h=-3x^2+12x+4x+1[/tex]Step 4. Combine like terms:
[tex]3g-h=-3x^2+16x+1[/tex]What we just found is (3g-h)(x):
[tex](3g-h)(x)=-3x^2+16x+1[/tex]Step 5. To find what we are asked for
[tex]\mleft(3g-h\mright)\mleft(-3\mright)[/tex]We need to evaluate the result from step 4, when x is equal to -3:
[tex](3g-h)(-3)=-3(-3)^2+16(-3)+1[/tex]Solving the operations:
[tex](3g-h)(-3)=-3(9)^{}-48+1[/tex][tex](3g-h)(-3)=-27^{}-48+1[/tex][tex](3g-h)(-3)=-74[/tex]Answer:
[tex](3g-h)(-3)=-74[/tex]The slope and y-intercept of the relation represented by the equation 12x-9y+12=0 are:
12x - 9y +12 =0
To find the slope and y intercept, we want to put the equation in slope intercept form
y = mx+b where m is the slope and b is the y intercept
Solve the equation for y
Add 9y to each side
12x - 9y+9y +12 =0+9y
12x+12 = 9y
Divide each side by 9
12x/9 +12/9 = 9y/9
4/3 x + 4/3 = y
Rewriting
y = 4/3x + 4/3
The slope is 4/3 and the y intercept is 4/3
The answer is 57.3 provided by my teacher, I need help with the work
Apply the angles sum property in the triangle ABC,
[tex]62+90+\angle ACB=180\Rightarrow\angle ACB=180-152=28^{}[/tex]Similarly, apply the angles sum property in triangle BCD,
[tex]20+90+\angle BCD=180\Rightarrow\angle BCD=180-110=70[/tex]From triangle ABC,
[tex]BC=AC\sin 62=30\sin 62\approx26.5[/tex]From triangle BDC,
[tex]BD=BC\cos 20=26.5\cos 20\approx24.9[/tex]Now, consider that,
[tex]\angle BDE+\angle BDC=180\Rightarrow\angle BDE+90=180\Rightarrow\angle BDE=90[/tex]So the triangle BDE is also a right triangle, and the trigonometric ratios are applicable.
Solve for 'x' as,
[tex]x=\tan ^{-1}(\frac{BD}{DE})=\tan ^{-1}(\frac{24.9}{16})=57.2764\approx57.3[/tex]Thus, the value of the angle 'x' is 57.3 degrees approximately.ang
Use an inequality to represent the corresponding Celsius temperature that is at or below 32° F.
C ≤ 0
Explanations:The given equation is:
[tex]F\text{ = }\frac{9}{5}C\text{ + 32}[/tex]Make C the subject of the equation
[tex]\begin{gathered} F\text{ - 32 = }\frac{9}{5}C \\ 9C\text{ = 5(F - 32)} \\ C\text{ = }\frac{5}{9}(F-32) \end{gathered}[/tex]At 32°F, substitute F = 32 into the equation above to get the corresponding temperature in °C
[tex]\begin{gathered} C\text{ = }\frac{5}{9}(32-32) \\ C\text{ = }\frac{5}{9}(0) \\ C\text{ = 0} \end{gathered}[/tex]The inequality representing the corresponding temperature that is at or below 32°F is C ≤ 0
Can you please solve the last question… number 3! Thanks!
Let us break the shape into two triangles and solve for the unknowns.
The first triangle is shown below:
We will use the Pythagorean Theorem defined to be:
[tex]\begin{gathered} c^2=a^2+b^2 \\ where\text{ c is the hypotenuse and a and b are the other two sides} \end{gathered}[/tex]Therefore, we can relate the sides of the triangles as shown below:
[tex]25^2=y^2+16^2[/tex]Solving, we have:
[tex]\begin{gathered} y^2=25^2-16^2 \\ y^2=625-256 \\ y^2=369 \\ y=\sqrt{369} \\ y=19.2 \end{gathered}[/tex]Hence, we can have the second triangle to be:
Applying the Pythagorean Theorem, we have:
[tex]22^2=x^2+19.2^2[/tex]Solving, we have:
[tex]\begin{gathered} 484=x^2+369 \\ x^2=484-369 \\ x^2=115 \\ x=\sqrt{115} \\ x=10.7 \end{gathered}[/tex]The values of the unknowns are:
[tex]\begin{gathered} x=10.7 \\ y=19.2 \end{gathered}[/tex]A Parks and Recreation department in a small city conducts a survey to determine what recreational activities for children it should offer. Of the 1200 respondents,400 parents wanted soccer offered625 parents wanted baseball/softball offered370 parents wanted tennis offered150 parents wanted soccer and tennis offered315 parents wanted soccer and baseball/softball offered230 parents wanted baseball/softball and tennis offered75 parents wanted all three sports offeredHow many parents didn’t want any of these sports offered?a) 155b) 75c) 0d) 425
There were 1200 respondents:
400 parents wanted soccer offered
370 parents wanted tennis offered
625 parents wanted baseball/softball offered
150 parents wanted soccer and tennis offered
315 parents wanted soccer and baseball/softball offered
230 parents wanted baseball/softball and tennis offered
75 parents wanted all three sports offered
Therefore:
We have to add all the options that were mixed with different sports:
150 + 315 + 230 = 695 - 75= 620 (We subtract 75 because it's already included in the other values as we can see in the diagram)
We have to add all the parents that chose only one sport:
400 + 370 + 625= 1395
We have to subtract 620 from 1395:
1395 - 620= 775 parents who chose any sport
Now:
1200 - 775 = 425 respondents who didn't want any.
Therefore, 425 parents didn't want any of the sports offered.
The answer is D) 425.
9. SAILING The sail on Milton's schooner is the shape of a 30°-60°-90°triangle. The length of the hypotenuse is 45 feet. Find the lengths of thelegs. Round to the nearest tenth.
The triangle is shown below:
Notice how this is an isosceles triangle.
We can find the lengths of the hypotenuse by using the trigonometric functions:
[tex]\sin \theta=\frac{\text{opp}}{\text{hyp}}[/tex]Then we have:
[tex]\begin{gathered} \sin 45=\frac{21}{hyp} \\ \text{hyp}=\frac{21}{\sin 45} \\ \text{hyp}=29.7 \end{gathered}[/tex]Therefore the hypotenuse is 29.7 ft.
How much water must be evaporated from 8 grams of a 30% antiseptic solution to produce a 40% solution?
Answer:
Step-by-step explanation:
8 grams of 30% --> 2.4 grams of AS For 2.4 grams to be 40% --> 6 grams of solution Evaporate 2 grams of water
In the triangle below, suppose that mZW=(x+4)º, mZX=(5x-4)°, and mLY= (4x)".Find the degree measure of each angle in the triangle.
Need answer if you could show work would be nice
use prowers and multiplication to write the equation whose value is 10 to the 11th power
if we have
(10^9)(10^2)
adds the exponents
10^(9+2)
10^11
If you have
10^18/ 10^7
subtract the exponents
10^(18-7)
10^11
If you have
(10^6)^2/10
First multiply the exponents
10^(6*2)/10
10^12/10
subtract exponents
10^(12-1)
10^11
factor completely5r^3-10r^2+3r-6
You have the following polynomial:
5r³ - 10r² + 3r - 6
In order to factorize the given polynomial, use synthetic division:
5 -10 3 -6 | 2
10 0 6
5 0 3 0
The remainder is zero in the previous division, then, r - 2 is a factor of the given polynomial, the other factor is formed with the coefficients of the division, just as follow:
5r³ - 10r² + 3r - 6 = (r - 2)(5r² + 3)
Hence, the factor are (r - 2)(5r² + 3)
Answer:(r-2) x (5r^2+3)
Step-by-step explanation:
In windy cold weather, the increased rate of heat loss makes the temperature feel colder than the actual temperature. To describe an equivalent temperature that more closely matches how it “feels,” weather reports often give a windchill index, WCI. The WCI is a function of both the temperature F(in degrees Fahrenheit) and the wind speed v (in miles per hour). For wind speeds v between 4 and 45 miles per hour, the WCI is given by the formula(FORMULA SHOWN IN PHOTO)A) What is the WCI for a temperature of 10 F in a wind of 20 miles per hour?B) A weather forecaster claims that a wind of 36 miles per hour has resulted in a WCI of -50 F. What is the actual temperature to the nearest degree?
Let's remember what the variables mean:
F= temperature (in Fahrenheit),
v= wind speed.
A) The formula "works" when the wind speed is between 4 and 45 miles per hour. The question asks for a wind speed of 20 miles per hour. Then, we can apply the formula. Here,
[tex]\begin{cases}F=10 \\ v=20\end{cases}[/tex]Then,
[tex]\begin{gathered} WCI(10,20)=91.4-\frac{(10.45+6.69\cdot\sqrt[]{20}-0.447\cdot20)(91.4-10)}{22}\approx\ldots \\ \ldots91.4-116.2857=-24.8857 \end{gathered}[/tex]Approximating, the answer is
[tex]-25F[/tex]B) This question is just about to find F in the provided equation after replacing the given v and WCI. Let's do that:
[tex]\begin{gathered} -50=91.4-\frac{(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F)}{22}, \\ -141.4=-\frac{(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F)}{22}, \\ -3110.8=-(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F), \\ 3110.8=(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F), \\ \frac{3110.8}{10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36}=91.4-F, \\ F=91.4-\frac{3110.8}{10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36}\approx1.2 \end{gathered}[/tex]Then, the actual temperature is
[tex]1F[/tex]determine if the following equations represent a linear function if so write it in standard form Ax+By=C9x+5y=102y+4=6x
9x + 5y = 10
is a linear equation because all variables are raised to exponent 1.
This equation is already written in standard form (A = 9, B = 5, C = 10)
2y + 4 = 6x
is a linear equation because all variables are raised to exponent 1.
Subtracting 2y at both sides:
2y + 4 - 2y= 6x - 2y
4 = 6x - 2y
or
6x - 2y = 4
which is in standard form (A = 6, B = -2, C = 4)
Can a triangle be formed with side lengths 17, 9, and 8? Explain.
Yes, because 17 + 9 > 8
Yes, because 17 + 8 < 9
No, because 9 + 8 > 17
No, because 8 + 9 = 17
Answer:
(d) No, because 8 + 9 = 17
Step-by-step explanation:
You want to know if side lengths 8, 9, and 17 can form a triangle.
Triangle inequalityThe triangle inequality requires the sum of the two short sides exceed the length of the longest side. For sides 8, 9, 17, this would require ...
8 + 9 > 17 . . . . . . . not true; no triangle can be formed
The sum is 8+9 = 17, a value that is not greater than 17. The triangle inequality is not satisfied. So, no triangle can be formed.
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What would the answer be?
Nvm, I got it wrong
Applying the definition of similar triangles, the measure of ∠DEF = 85°.
What are Similar Triangles?If two triangles are similar, then their corresponding angles are all equal in measure to each other.
In the image given, since E and F are the midpoint of both sides of triangle BCD, then it follows that triangles BCD and EFD are similar triangles.
Therefore, ∠DBC ≅ ∠DEF
m∠DBC = m∠DEF
Substitute
4x + 53 = -6x + 133
4x + 6x = -53 + 133
10x = 80
10x/10 = 80/10 [division property of equality]
x = 8
Measure of ∠DEF = -6x + 133 = -6(8) + 133
Measure of ∠DEF = 85°
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You go to the pet store with $25. You decide to buy 2 fish for $3.69 each and fish foos for $4.19. Rounded tanks are $11.48 square-shaped tanks are $14.89. Estimate your total cost to find which tank you can can buy. About how much money will you have left?
Answer: you will only have enough money for the rounded tank, after buying everything you will have 9 cents left
Step-by-step explanation: two $3.69 fish, $4.19 fish food. 2x3.69=7.38+4.19=11.57
25-11.57=13.43
13.43+11.48=24.91
25-24.91=0.09
Surface are of the wood cube precision =0.00The weight of the woo cube precision =0.00 The volume was 42.87 in3
Given:
The volume of the cube is 42.87 cubic inches.
The volume of a cube is given as,
[tex]\begin{gathered} V=s^3 \\ 42.87=s^3 \\ \Rightarrow s=3.5 \end{gathered}[/tex]The surface area of a cube is,
[tex]\begin{gathered} SA=6s^2 \\ SA=6\cdot(3.5)^2 \\ SA=73.5 \end{gathered}[/tex]Answer: the surface area is 73.5 square inches ( approximately)
Yasmin went to the store and bought 3 and 1/2 pounds of ground beef for 11:20 how much do the ground beef cost per pound
Yasmin bought 3 1/2 pounds of ground beef, we can express the amount that she bought as a fraction like this:
[tex]3\frac{1}{2}=\frac{3\times2+1}{2}=\frac{6+1}{2}=\frac{7}{2}[/tex]Since she bought it for $11.2, if we divide the cost by the amount that she purchased, we get the cost per pound, like this:
[tex]\frac{11.2}{\frac{7}{2}}[/tex]To divide by a fraction, we just have to invert its numerator and denominator:
[tex]\frac{11.2}{\frac{7}{2}}=11.2\times\frac{2}{7}=\frac{22.4}{7}=3.2[/tex]Then, the cost per pound equals $3.2