Given system of equations have no solutions that is nonzero.
We have been given the system of equations formed by x − y = 4 and ay - ax + 4a = 0
We need to find the number of non-zero solutions the system have for a nonzero number 'a'
x - y = 4 ............(1)
ay - ax + 4a = 0 ............(2)
From equation (1),
x = 4 + y
Substitute x = 4 + y in equation (2),
ay - a(4 + y) + 4a = 0
ay - 4a - ay + 4a = 0
-4a + 4a = 0
0 = 0
This means that the system have no solutions that is nonzero.
Therefore, given system of equations have no solutions that is nonzero.
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Which equation is equivalent to: 3r=78+14 ?A. −3r=−78+14B. 3r−14=78C. 3r=78−14D. −3r=78−14
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The general form of an ellipse is 15x2+4y2+30x−16y−29=0.
What is the standard form of the ellipse?
The equation of ellipse in standard form is (x + 1)² / 4 + (y - 2)² / 15 = 1.
How to determine the standard form of the ellipseIn this problem we find the equation of an ellipse in general form, whose standard form can be found by algebra properties. The general form of the equation of an ellipse centered at (h, k) is introduced below:
(x - h)² / a² + (y - k)² / b² = 1
Where:
a, b - Lengths of the semiaxes.(h, k) - Coordinates of the center.The complete procedure is now presented:
15 · x² + 4 · y² + 30 · x - 16 · y - 29 = 0
(15 · x² + 30 · x) + (4 · y² - 16 · y) = 29
15 · (x² + 2 · x) + 4 · (y² - 4 · y) = 29
15 · (x² + 2 · x + 1) + 4 · (y² - 4 · y + 4) = 29 + 15 · 1 + 4 · 4
15 · (x + 1)² + 4 · (y - 2)² = 29 + 15 + 16
15 · (x + 1)² + 4 · (y - 2)² = 60
(x + 1)² / 4 + (y - 2)² / 15 = 1
The equation of ellipse is (x + 1)² / 4 + (y - 2)² / 15 = 1.
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Solve the missing elements for each problem. Use 3.14 for π. Area = πr^2; C=π D
Given,
Diameter = 32 cm
Radius
We know the radius is half of the diameter. Thus,
[tex]\begin{gathered} r=\frac{32}{2} \\ r=16 \end{gathered}[/tex]Radius 16 cm
Circumference
The formula is:
[tex]C=\pi D[/tex]Where
D is the diameter
So,
[tex]\begin{gathered} C=\pi D \\ C=(3.14)(32) \\ C=100.48 \end{gathered}[/tex]Circumference = 100.48 cm
Area
The formula is:
[tex]A=\pi r^2[/tex]Where
r is the radius
So,
[tex]\begin{gathered} A=\pi r^2 \\ A=(3.14)(16)^2 \\ A=(3.14)(256) \\ A=803.84 \end{gathered}[/tex]Area = 803.84 sq. cm.
what is the solution set for the inequality
A. x ≤ -5
B. x ≤ 5
C. x ≤ 1
d. x ≤ -14
The solution set to the inequality, 4x + 12 ≤ -8, is determined as: A. x ≤ -5.
How to Find the Solution Set of an Inequality?The solution set is the value of x that would make an inequality statement true. To find the value of x, solve as you would solve a normal equation.
Using the key to the given model in the diagram, we can write the inequality as follows:
On the left, we would have, 4x + 12.
On the right, we would have, -8.
The inequality would be expressed as:
4x + 12 ≤ -8
Solve for x
4x + 12 - 12 ≤ -8 - 12 [subtraction property of equality]
4x ≤ -20
4x/4 ≤ -20/4
x ≤ -5
The solution set is: A. x ≤ -5.
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F(x)=1/x g (x)=x-4 can you evaluate (golf)(0)? Explain why or why not.
If f(x) = 1/x and g(x) = x-4 , then (gof)(0) cannot be evaluated as the function becomes not defined .
In the question
it is given that the functions
f(x) = 1/x
and g(x) = x-4
to find g=(gof)(x) ,
(gof)(x) = g(f(x))
= g(1/x) ... because f(x) = 1/x
= 1/x - 4 ....because g(x) = x-4
On simplifying further , we get
= (1-4x)/4x
On substituting x=0 , we get
gof(0) = (1-0)/4(0)
= 1/0
which is not defined , hence cannot be evaluated.
Therefore , if f(x) = 1/x and g(x) = x-4 , then (gof)(0) cannot be evaluated as the function becomes not defined .
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Your brother is buying textbooks for college. He has to buy 3 math textbooks and 2 science textbooks. The total cost of his textbooks is $487. Write a linear equation to represent the cost of his textbooks.
Let's define the following variables,
x: cost of a math textbook
y: cost of a science textbook
He has to buy 3 math textbooks and 2 science textbooks, that is,
Total cost = 3x + 2y
The total cost of his textbooks is $487, then the linear equation is,
487 = 3x + 2y
when graphed on a coordinate plane,Bumby Avenue can be represented by the equation y=-4x-7. primrose can be represented by the equation 8x+2y=17. Are these streets parallel ?
Answer:
The lines are not parallel because their slopes are opposite reciprocals.
Explanation:
The lines:
[tex]\begin{gathered} y=-4x-7 \\ \text{and} \\ 8x+2y=17 \end{gathered}[/tex]are not parallel because their slopes are not the same
Note:
Two straight lines are said to be parallel when their slopes are the same, and have different y-intercepts.
this lettuce i have is 25 calories per serving. serving size is 85 grams. i had 27 grams. how many calories would that be??
Answer:
7.94117647059 Calories
Step-by-step explanation:
25 Calories = 85 Grams
25/85 = 0.294117647059
0.294117647059 = 1 Gram
0.294117647059 times 27 = 7.94117647059
Fiona is making a banner in the shape of a triangle for a school project. She graphs the banner on a coordinate plane with vertices at P(0, 4) , Q(2, 8) , and R(−3, 6) . She wants to reflect the banner over the line x=1. Identify the image of the banner reflected in the line x=1.
The coordinates of the banner after the reflection across x = 1 is P'(2, 4), Q'(0, 8), and R'(5, 6)
How to determine the coordinates of the banner after the reflection?From the question, the coordinates are given as
P(0, 4), Q(2, 8), and R(−3, 6)
The line of reflection is given as
x = 1
The rule of reflection across the line x = 1 is represented as
(x, y) = (-x + 2, y)
When the above rule is applied, we have
P'(2, 4), Q'(0, 8), and R'(5, 6)
This means that the coordinate of the image are P'(2, 4), Q'(0, 8), and R'(5, 6)
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Use the times and corresponding closing prices of the stock to create coordinate pairs. Let X represent the number of weeks since the first at a point, and Y represent the closing price of each time. So, X equals zero represents the data point from five years ago. There are 52 weeks in a year, and you can write the time for each closing price recorded in terms of weeks that have passed since five years ago, when X equals zero. Fill in the table to represent your data as coordinate pairs
Combining both tables we get:
CASSANDRA WENT FOR A JO9.SHE RAN AT A PACE OF 7.3 MILESPER HOUR. IF SHE RAN FOR 0.75HOURS, HOW FAR DID CASSANDRARUN?
We can use one simple formula, that is d=vt
d=distance
v=pace
t=time
So,
d=(7.3miles per hour)(0.75 hours)=5.475 miles
Your cousin is building a Sandbox for his daughter how much sand will he need to fill the Box? Explain. How much paint will he need to paint all six surface of the sandbox? Explain. 1ft 4ft 6ft not answer choices
Since the image is a rectangular prism
The volume of the box can be obtained by using the formula:
Volume = l x b x h
The box has a dimension of 1ft x 4ft x 6ft
The volume of the box = 1 x 4 x 6 = 24 cubic feet
Therefore, the volume of sand needed to fill the box will be = 24 cubic feet of sand
The surface area of the box can be obtained using the formula:
2(lb + lh + bh)
= 2(1x4 + 1x6 + 4x6)
=2(4 + 6 + 24)
=2 (34)
= 68 square feet
Therefore a total surface area of 68 square feet needs to be painted
can you help me solve this in expanded form. 156 X 687 = ?
Given data:
The given expression is 156x687.
The given expression can be written as,
[tex]\begin{gathered} (100+50+6)(600+80+7)=60000+8000+700+30000+4000+350+3600+480+42 \\ =107172 \end{gathered}[/tex]Thus, the value of the given expression is 107172.
The endpoints of diameter in a circle form an angle with point C. What is the measure of angle BCD?
Answer:
90 degrees
Explanation:
Please note that whenever a triangle is inscribed in a circle and one side of the triangle is a diameter of the circle as in the given figure, then the angle that is opposite the diameter of the circle is a right angle;
[tex]\therefore\angle BCD=90^{\circ}[/tex]Solve the following/3x=7-3/x
Value of x is 8.46 for equation 3x=7-3/x
What is Equation?Two or more expressions with an Equal sign is called as Equation
The given equation is 3x=7-3/x
3x+3/x=7
3x²+3=7x
3x²-7x+3=0
Use quadratic equation formula
a=3, b=-7, c=3
x=-b±√b²-4ac/2a
x=7±√49-4(3)(3)/2(3)
x=7±√49-36/6
x=7±√13/6
x=7±√2.16
x=7+1.46
x=8.46
Hence value of x is 8.46 for equation 3x=7-3/x
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For the following set of data, find the number of data within 2 population standard deviations of the mean.28, 65, 114, 74, 68, 75, 70, 69, 64
To determine the data that is within 2 population standard deviations of the mean, let's calculate the mean first.
To determine the mean, let's add all the data and divide the result by the total number of data.
[tex]28+65+114+74+68+75+70+69+64=627[/tex][tex]627\div9=69.66667[/tex]The mean is 69.66667.
Let's now calculate the standard deviation. Here are the steps:
1. Subtract the mean from each data, then square the result.
[tex]\begin{gathered} 28-69.66667=(-41.66667)^2=1,736.1114 \\ 65-69.66667=(-4.66667)^2=21.7778 \end{gathered}[/tex][tex]\begin{gathered} 114-69.66667=(44.33333)^2=1,965.4441 \\ 74-69.66667=(4.33333)^2=18.7777 \end{gathered}[/tex][tex]\begin{gathered} 68-69.66667=(-1.66667)^2=2.7778 \\ 75-69.66667=(5.33333)^2=28.4444 \end{gathered}[/tex][tex]\begin{gathered} 70-69.66667=(0.33333)^2=0.1111 \\ 69-69.66667=(-0.66667)^2=0.4444 \\ 64-69.66667=(-5.66667)^2=32.1111 \end{gathered}[/tex]2. Add the results in step 1.
[tex]1,736.1114+21.7778+1,965.4441+18.7777+2.7778=3,744.8888[/tex][tex]28.4444+0.1111+0.4444+32.1111=61.111[/tex][tex]3,744.8888+61.111=3,805.9998[/tex]The sum is 3, 805.9998.
3. Divide the sum by the total number of data.
[tex]3,805.9998\div9=422.8889[/tex]4. Square root the result in step 3.
[tex]\sqrt{422.8889}\approx20.56[/tex]The standard deviation is approximately 20.56.
So, the data that are within 2 population standard deviations of the mean are between:
[tex]\begin{gathered} 69.67-(2)(20.56)=28.55\approx29 \\ 69.67+(2)(20.56)=110.79\approx111 \end{gathered}[/tex]The data that are within 2 population standard deviations of the mean are between 29 and 111. Based on the given data, the data that are between 29 and 111 are the following: 64, 65, 68, 69, 70, 74, and 75. There are 7 data that are within 2 population standard deviations of the mean.
Find the distance between:(4,-9) and(-8,0)Round your answer to the nearest hundredth.
The distance between 2 points (x1, y1) and (x2, y2) is calculated as:
[tex]\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]So, if we replace (x1, y1) by (4, -9) and (x2, y2) by (-8,0), we get:
[tex]\begin{gathered} \text{distance}=\sqrt{(-8-4)^2+(0-(-9))^2} \\ \text{distance}=\sqrt{(-12)^2+(9)^2} \\ \text{distance}=\sqrt{144+81} \\ \text{distance}=\sqrt{225} \\ \text{distance}=15 \end{gathered}[/tex]Answer: the distance is 15
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The rectangle has a width of 7 yards and a length of 42 yards.
How to find the length and width?For a rectangle of length L and width W, the perimeter is:
P = 2*(L + W)
Here we know that:
L = 6*W
P = 98yd = 2*(L + W)
Then we have the system of equations:
L = 6*W
98yd = 2*(L + W)
If we substitute the first equation into the second one, we get:
98yd = 2*(6*W + W)
98yd = 14*W
98yd/14 = W = 7yd
So the width is 7 yards, and the length is 6 times that, so:
L = 6*7yd = 42yd
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Section 5.2-10. Solve the following system of equations by substitution or elimination. Enter your answer as (x,y).-2x+3y = 15-x-3y = 12
1)-2x2)
[tex]-2x+3y-(-2x-6y)=15-2\times12\Rightarrow-2x+3y+2x+6y=15-24\Rightarrow9y=-9\Rightarrow y=-1[/tex]y=-1 implies
[tex]-2x+3\times(-1)=15\Rightarrow-2x-3=15\Rightarrow-2x=18\Rightarrow x=-9[/tex]Hence the solution is
[tex](-9,-1)[/tex]The standard normal curve is grafted below. Shade the region under the standard normal curve to the left of x=1.00Use the table to find the area under the standard normal curve to the left of x=1.00
Explanation
Part A
The shaded area under the standard normal curve to the left of z=1.00 can be seen below.
Part B
Using the z table, the area under the standard normal curve to the left of z.=1.00 is
Answer: 0.8413
in order to clean her aquarium Stephanie much remove half of the water the garden measures 30 inches long 16 inches wide and 12 inches deep the aquarium is currently completely full with volume of water in cubic inches must Stephanie remove?
Hello!
30in long (length)
16in wide (width)
12in deep (height)
First, we have to calculate the volume when the aquarium is full of water, using the formula:
[tex]undefined[/tex]Can I Plss get some help I got stuck I don’t know how to find x
Using Sine of angles to evaluate for x
The formula is,
[tex]sin\theta=\frac{Opposite}{Hypotenuse}[/tex]Given:
[tex]\begin{gathered} Opposite=x \\ Hypotenuse=19 \\ \theta=21^0 \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} sin21^0=\frac{x}{19} \\ \therefore x=19\times sin21^0 \end{gathered}[/tex]Simplify
[tex]x=6.80899\approx6.81\text{ \lparen2 decimal places\rparen}[/tex]Hence,
[tex]x=6.81[/tex]A bookstore sells a college algebra book for $90. If the bookstore makes a profit of 25% on each sale,what does the bookstore pay the publisher for each book?
Okay, here we have this:
Considering the provided information, we obtain that:
The total price = Commission of the bookstore + Payment to the publisher
Replacing:
$90=$90(0.25)+Payment to the publisher
Payment to the publisher=$90-$90(0.25)
Payment to the publisher=$90-$22.5
Payment to the publisher=$67.5
Finally we obtain that the bookstore pay $67.5 to the publisher for each book.
How would these look graphed ? Look at image attached .
These are two lines intersected ,in one point
One is positive inclined, the other negative.
Then now GRAPH
THEN BOTH LINES INTERSECT AT
Triangle KLM has KL = 28, KM = 28, and LM = 21. What is the area of the triangle?The area of AKLM is about(Simplify your answer. Round to one decimal place as needed.)
Area = (b * h)/2, b = 21 bu we don't know h, s we have to calculate it
To calculate the height "h" we can use pythagoras with a triangle rectangle with base = 21/2 = 10.5 and hypothenuse = 28, so the height "h" is:
28² = h² + 10.5² ==> h² = 28² - 10.5² = 784 - 110.25 = 673.75
h² = 673.75
h = 25.96
Now that we have the height, the Area of the triangle = (b * h)/2 = (21 * 25.96)/2 = 272.5
Answer:
272.5
a regular octagon has an area of 49 m 2 . find the scale factor of this octagon to a similar octagon with an area of 100 m 2
Given,
The area of the regular octagon is 49 square metre.
The area of the another regular octagon is 100 square metre.
[tex]\begin{gathered} \text{Scaling factor=}\frac{\sqrt{\text{area of regular polygon}}}{\sqrt[]{\text{area of another regular plogon}}\text{ }} \\ \text{Scaling factor=}\frac{\sqrt[]{\text{4}9}}{\sqrt[]{\text{1}00}\text{ }} \\ \text{Scaling factor=}\frac{7}{10\text{ }} \end{gathered}[/tex]Here, the scaling factor of the regualar octagon is 7:10
Hence, the scaling factor is 7:10.
For an arc length s, area of sector A, and central angle θ of a circle of radius r, find the indicated quantity for the given value. r= 6.45 in, θ= 5 pi\6, s=?
Calculate the arc length by using the following formula:
[tex]s=r\theta[/tex]Replace the values of r and θ and simplify:
[tex]\begin{gathered} s=(6.45in)(5\frac{\pi}{6})=(6.45)(\frac{5}{6})(3.14) \\ s=16.8775in \end{gathered}[/tex]Hence, the arc length is 16.8775 in
the ratio of the length to the width of a rectangular hall is 5:3. if the width is 1500cm, find the lenght.
Step-by-step explanation:
The ratio of the length to the width that is- length:width = 5:3
Take x as a common value,
5x= length
3x= width
Width of the rectangle= 1500 cm
3x= 1500 cm
x= 1500/3
x= 500 cm
Length of the rectangle= 5x
x=500 cm
Length= 5*500
=2500 cm
Length of the rectangle= 2500 cm
4. A teacher can only make 3000 copies in a month. If a teacher-has-made 2700 copies so far this month, what percentage of her copies has she used?
In order to determine the percentage, let x as the percentage. Then, you can write:
[tex]\frac{x}{100}\cdot3000=2700[/tex]factor x/100 is the percentage in decimal form. The product of this factor and 3000 equals 2700.
Solve for x and simplify:
[tex]x=\frac{2700}{3000}\cdot100=90[/tex]Hence, teacher has used 90% of the copies.
Hello, I had a question on how to find the leading coefficient and the degree.
Given:
given polynomial is
[tex]23v^5-2v+4v^8-18v^4[/tex]Find:
we have to find the leading coefficient and degree of the polynomial.
Explanation:
The lewading coefficient is the coefficient of highest power term of the polynomial.
Highest power of v is 8 and its coefficient is 4.
Therfore, leading coefficient is 4.
and the degree of the polynomial is equal to the highest power of v in the polynomial, which is 8.
Therefore, the leading coefficient of polynomial is 4 and degree is 8.