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Answer:

y=90(0.5)^x

The y-intercept is (0,90), so when x=0, y should be 90.

By replacing x=0 in all equations you get;

1)

y=90(2)^0 =90 accepted

2)

y=2(90)^0 =2 rejected

3)

y=0.5(90)^0 =0.5 rejected

4)

y=90(0.5)^0 =90 accepted

Since this is an exponential decrease,

y=90(2)^x is rejected.

[tex]\text{R.H.S}\\\\=4\sin 45^{\circ} \cos 15^{\circ} \\\\=4\sin 45^{\circ} \cos (45^{\circ}-30^{\circ})\\ \\=4 \sin 45^{\circ} \left(\cos 45^{\circ} \cos 30^{\circ} +\sin 45^{\circ} \sin 30^{\circ} \right)\\\\=4 \left(\sin 45^{\circ} \cos 45^{\circ} \cos 30^{\circ} +\sin^2 45^{\circ} \sin 30^{\circ} \right)\\[/tex]

[tex]=4\left[\dfrac 1{\sqrt 2} \cdot \dfrac 1{\sqrt 2} \cdot \dfrac{\sqrt 3}2 +\left( \dfrac 1{\sqrt 2} \right)^2 \cdot \dfrac 12 \right]\\ \\=4\left(\dfrac{\sqrt 3}4 +\dfrac 14\right)\\\\=4\cdot \dfrac 14 \left(1+\sqrt 3\right)\\\\=1+\sqrt 3\\\\=\text{L.H.S}[/tex]

Step-by-step explanation:

The Volume of a Sphere ( One piece of candy ) = 4/3πr³

r = diameter/2 = 9/2 = 4.5 cm

Volume = 4/3(3.14)(4.5³)

V = 381.51 cm³

the box contains 10 pieces so the box contains :

381.51 x 10 = 3815.1 cm³ of chocolate

Answer:

3 US galloons

Answer:

population

Step-by-step explanation:

Answer: 1/5 6/24 1/3 2/4 3/4

Step-by-step explanation: bigger the denominator smaller the value.

Answer:

20 lb of 31 cent candy, 30 lb of 46 cent candy

Step-by-step explanation:

sorry don't have one

here is a smiley face for you

:D

The amount of each type of candy (in pounds) that are mixed for the considered condition is: 20 pounds of first type of candy and 30 pounds of second type of candy.

You can represent the unknown amounts by the use of variables. Follow whatever the description is and convert it one by one mathematically. For example if it is asked to increase some item by 4 , then you can add 4 in that item to increase it by 4. If something is for example, doubled, then you can multiply that thing by 2 and so on methods can be used to convert description to mathematical expressions.

For this case, we can assume the separate weights of each type of candies to be represented by variables.

Let we have:

x pounds of first type of candy is mixed with y pounds of second candy.

The weight of the mixture = x + y = 50 pounds

The cost of the mixture = 40 cents per pound  = [tex]40 \times 50 = 2000[/tex] cents

Cost of the mixture = cost of x pounds of first candy+ cost of y pounds of second candy = [tex]31x + 46y[/tex]
Thus, we get [tex]31x + 46y =2000[/tex]

Therefore, we got a system of two linear equations as:

[tex]x +y = 50\\31x + 46y =2000[/tex]

From first equation, we get:

[tex]x = 50 -y[/tex]

Substituting this value in second equation, we get:

[tex]31(50-y) + 46y =2000\\1550 + 15y = 2000\\15y = 450\\\\y = 450/15 = 30[/tex]

Putting this value of y in equation for x, we get:

[tex]x = 50-y = 50-30 = 20[/tex]

Thus, the amount of each type of candy (in pounds) that are mixed for the considered condition is: 20 pounds of first type of candy and 30 pounds of second type of candy.

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Answer:

hello

( 20 - 2 x ) ( 18 - 2 x )

= 360 - 40 x - 36 x + 4 x²

= 4 x² - 76 x + 360

Step-by-step explanation:

Answer:

1,000

Step-by-step explanation:

Answer:

80

Step-by-step explanation:

area of parallelogram is b(vh)

Answer:

1360 seats sold

Step-by-step explanation:

total seats = 1600

total seats sold = 85% of 1600

= 85/100 × 1600

= 85 × 16

= 1,360 seats sold

Answer:

£78

Step-by-step explanation:

65 divided by 5= £13

13+65=£78

Answer:

£78

Step-by-step explanation:

Let the total value be T.

Considering the situation described, the appropriate hypotheses set is given by:

[tex]H_0: p = 0.26[/tex]

[tex]H_1: p > 0.26[/tex]

At the null hypotheses, we test if there is not enough evidence that the percentage has increased, that is, it still is of 26%, hence:

[tex]H_0: p = 0.26[/tex]

At the alternative hypotheses, we test if there is evidence that it has increased, hence:

[tex]H_1: p > 0.26[/tex]

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mTQR = 87°

HOPE IT HELPS...!!!

Answer:

[tex] \sf m \angle TQR = 87 \degree[/tex]

Step-by-step explanation:

Given:

∆QRT, ∆RSQ and ∆ QST are scalene traingle.

[tex]m \angle QRS = 37 \degree \\ m \angle TQS = 34 \degree \\ m \angle QST = 90 \degree[/tex]

To find:

[tex]m \angle TQR = ?[/tex]

Solution:

In ∆ QTS,

[tex] \sf \: m \angle TQS + m \angle QTS +m \angle QST = 180 \degree \: \\ \sf (properties \: of \: triangle)[/tex]

[tex] \sf \: 34 + m \angle QTS +90 = 180 \degree \: \\ \sf m \angle QTS = 180 - 90 - 34 \\ \sf \: m \angle QTS = 180 - 124 \\ \sf \: m \angle QTS = 56 \degree[/tex]

Now in ∆QTR,

[tex] \sf \: m \angle QTR + m \angle TQR + m \angle TRQ \: = 180 \degree \\ \sf (properties \: of \: triangle)[/tex]

[tex] \sf \: 56 + m \angle TQR + 37 \: = 180 \degree \\ \sf m \angle TQR = 180 - 56 - 37 \\ \sf m \angle TQR = 87 \degree[/tex]

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Answer:

A

First prove Triangle ABC is congruent to Triangle CDA, and then state AD and BC are corresponding sides of the triangles.

Hope this helps!

Please Mark Brainleast!!!

The sequence to prove that AD = BC is; First prove ABC is similar to CDA, and then state AD and BC are opposite sides of the parallelograms.

From the question, we see that the parallelogram ABCD has a diagonal AC. Thus; AB║ CD and AD║BC.

Now, the parallelogram is divided into two triangles ΔABC and ΔADC by its diagonal AC.

Thus;

∠ACB = ∠DAC  and ∠CAB = ∠ACD  because they are alternate interior angles.

Also, AC = AC  (Reflexive property)

Thus; by ASA congruence postulate, we can say that; ΔABC ≅  ΔADC

Also, by corresponding sides of the congruent triangles are congruent we can say that AD = BC.

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Answer:

(x – 7)² – 9

Step-by-step explanation:

I hope this helps

Answer:

-1/2

Step-by-step explanation:

2/5 * - 3/7 - 1/14 - 3/7 * 3/5

= - 6/35 - 1/14 - 9/35

= - (6*2)/(35*2) - (1*5)/(15*5) - (9*2)/(35*2)

= - 12/70 - 5/70 - 18/70

= (-12 - 5 - 18)/70

=- 35/70

= - 1/2

Answer:

4

Step-by-step explanation:

We can see that 9 households are represented in the data, 4 of them are zeros. This means 4 households in Cam's neighborhood have 0 children.

Answer:

x^2 - 2001 = 0

Step-by-step explanation:

I guess you mean √2001 and -√2001

The product of the 2 solutions = -2001 and the sum is 0.

The required equation  is x^2 - 2001 = 0

Answer:

1.3×10^1

Step-by-step explanation:

For this, you need to use scientific notation

First, let's round up the decimal points

135.99 = ~ 136

1.3 ×.10^1

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after approximately 137 minutes, there will be 10 bacteria remaining.

To find the time at which there will be 10 bacteria remaining, we can use the half-life model for exponential decay, which is given by:

N(t) = N₀ * [tex](1/2)^{(t / t_{half})[/tex]

where:

N(t) = the number of bacteria at time t

N₀ = the initial number of bacteria

t = time (in minutes)

t_half = the half-life of the bacteria

We are given that the initial number of bacteria (N₀) is 500 and after 17 minutes there are 150 bacteria (N(17) = 150).

Let's use this information to find the value of k (the decay constant):

150 = 500 * [tex](1/2)^{(17 / t_{half})[/tex]

Divide both sides by 500:

[tex](1/2)^{(17 / t_{half})[/tex] = 150 / 500

[tex](1/2)^{(17 / t_{half})[/tex] = 0.3

Now, we can write the equation as an exponent:

17 / t_half = log base (1/2) of 0.3

Using logarithms:

t_half = 17 / log base (1/2) of 0.3

Now, calculate t_half:

t_half = 17 / log base 2 of 0.3 ≈ 38.555

Now, we can find the time at which there will be 10 bacteria remaining (N(t) = 10):

10 = 500 * [tex](1/2)^{(t / 38.555)[/tex]

Divide both sides by 500:

[tex](1/2)^{(t / 38.555)[/tex] = 10 / 500

[tex](1/2)^{(t / 38.555)[/tex] = 0.02

Write the equation as an exponent:

t / 38.555 = log base (1/2) of 0.02

Using logarithms:

t = 38.555 * log base 2 of 0.02 ≈ 137.08

Rounding to the nearest whole number:

t ≈ 137

So, after approximately 137 minutes, there will be 10 bacteria remaining.

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Answer:

19000

Step-by-step explanation:

1 hour = 50 miles

2 hour = 100 miles

50×380 = 19000

Answer:

Well, i would just say you just had to do your best in everything that you do. No pressure , no stress don't overwork yourself and simply take action. Then you will have a chance to get what you want because you tried your best to get to it. Although sometimes we don't get what we want, atleast we'll be closer. As they say, 'Aim for the sun to reach the moon"

Step-by-step explanation:

You're welcome

in 20 days 4 half life will be completed

so n = 4

formula :-

[tex] \frac{ A\degree}{ {2}^{n} } = \frac{9056}{ {2}^{4} } = \frac{9056}{16} = 566grams[/tex]

The radioactive kind of promethium will have 566 grams left after 20 days.

Half-life is the time required for a quantity to reduce to half of its initial value.

Given, A radioactive kind of promethium with 9,056 grams and a half-life is 5 days.

∴ In the next 5 days, it'll have (9056/2) = 4528 grams.

Next 5 days it'll have (4528/2) = 2269 grams.

As it has a half-life in 5 days and 20 is multiple (4×5).

After 20 days the radioactive material will have (9056/2⁴) grams.

= 566 grams left.

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Answer:

f(x) = 2x3 + 4

Stretch vertically be a factor of 2:

f(x) = 2*2x3 + 4 = 4x3 + 4

Shift 5 units to the left:

f(x) = 4(x+5)3 + 4

The 39th term of the sequence is 200.

Arithmetic sequence or arithmetic progression is the set of numbers for which the difference of any two consecutive numbers in that set will be a constant.

The constant term which is the difference of the two consecutive numbers is called the common difference and is usually denoted by 'd'.

Arithmetic sequence is usually represented as

a, a + d, a + 2d, ........, a + (n - 1)d

The nth term of an arithmetic sequence, aₙ is given by the formula,

aₙ = a + (n - 1) d

where, a is the first term and d is the common difference.

The given sequence is 10, 15, 20, ......

a = 10

d = 15 - 10 = 5

Here, we have to find the 39th term of the sequence.

a₃₉ = 10 + (39 - 1) 5

     = 10 + (38 × 5)

     = 10 + 190

     = 200

Hence, the 39th term of the progression 10, 15, 20,..... is 200.

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Answer:

x = 49

Step-by-step explanation:

corner point b = 41

d = right angle ( 90 )

90 + 41 = 131

180- 131 = 49

x = 49

p.s

im not sure if d is a right angle if it is not please inform me.