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liraira [26]
3 years ago
14

I need help i am very confused

Mathematics
1 answer:
viva [34]3 years ago
5 0

Answer:

its not proportional

Step-by-step explanation:

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Solve 9y-(2y-3) =5(y-2) +2y​
EastWind [94]

Answer: it’s false there is no solution

Step-by-step explanation:

7 0
3 years ago
Use the following function rule to find f(3).<br> f(x) =4^x -5<br> F(3) =
BigorU [14]

Answer:

59

Step-by-step explanation:

To find f(3), substitute 3 for x:

f(3)=4^3-5\\f(3)=64-5\\f(3)=59

7 0
2 years ago
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∆ABC is similar to ∆DEF. The ratio of the perimeter of ∆ABC to the perimeter of ∆DEF is 1 : 10. The longest side of ∆DEF measure
nikdorinn [45]
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\bf \cfrac{\triangle ABC}{\triangle DE F}\qquad \cfrac{longest\ side}{longest\ side}\quad \cfrac{1}{10}=\cfrac{40}{s}\implies s=\cfrac{10\cdot 40}{1}\\\\&#10;-------------------------------\\\\&#10;\cfrac{\triangle ABC}{\triangle DE F}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}\implies \cfrac{1}{10}=\cfrac{\sqrt{A_1}}{\sqrt{A_2}}\implies \cfrac{1}{10}=\sqrt{\cfrac{A_1}{A_2}}&#10;\\\\\\&#10;\left( \cfrac{1}{10} \right)^2=\cfrac{A_1}{A_2}\cfrac{1^2}{10^2}=\cfrac{A_1}{A_2}\implies \cfrac{1}{100}=\cfrac{A_1}{A_2}
8 0
2 years ago
Read 2 more answers
Write a quadratic equation having the given solutions -8
Anna71 [15]

Answer:

y=x^2+16x+64

Step-by-step explanation:

A quadratic needs 2 solutions, but since you only have 1, I am to assume that x = -8 twice (or multiplicity of 2).  If x = -8, then in factor form, it's (x + 8).  Muliplying that by itself twice:

(x + 8)(x + 8):

x^2+8x+8x+64=0

Simplifying gives you:

x^2+16x+64=y

6 0
3 years ago
A sample of bacteria experienced 5% daily growth compounded continuously for 14 days. At the end of the 14 days, the sample had
Anton [14]

Answer:

sample size at the beginning was 40 bacteria.

Step-by-step explanation:

Given that a sample of bacteria experienced 5% daily growth compounded continuously for 14 days

Let C be the initial amount of bacteria.

C grows continuously compounding at the rate of 5%

The equation representing the number of bacteria at time t days would be

P(t) = Ce^{0.05t}

Now we have t = 14 days and P = 80.

Substitute to get

80=Ce^{0.05(14)} \\C=80e^{-0.05(14)}\\C=39.72682

Approximately 40 bacteria were present at the beginning.

6 0
3 years ago
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