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4 years ago

5

Need Help Since Schoo is Closed

2 answers:

lana [24]4 years ago

5 0

Answer:

32

Step-by-step explanation:

If a=5 and five to the second power is 25 then all you have to do is add 7+25=32

Hope this helps!!

marishachu [46]4 years ago

4 0

Answer:

32

Step-by-step explanation:

let a = 5

a² + 7 | Original equation

5² + 7 | Substitution of a as 5

25 + 7 | 5 to the power of 2 is 25, as 5*5 = 25.

32 | 25 + 7 = 32

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Which shapes could this hexagon be decomposed into to find its area? Choose all that apply (3 points).

Scilla [17]

You can do:

-two trapezoids
-six triangles
-two parallelograms

You can't put it into one rectangle, that will definitely not work.

As you can see from the attachment, triangles work, trapezoid as well, and two parallelograms. The one rectangle would not work.

I hope this helps!
~kaikers

3 0

3 years ago

Identify the radius of the circle whose equation is x^2 + y^2 = 36.

pantera1 [17]

Answer:

radius = 6

Step-by-step explanation:

The equation of a circle centred at the origin is

x² + y² = r² ←r is the radius

x² + y² = 36 ← is in this form

with r² = 36 ⇒ r = $\sqrt{36}$ = 6

3 0

4 years ago

10. For the equation 3x – 4 = 8, which

irga5000 [103]

Answer:

x = 4

Step-by-step explanation:

We can find the answer to this by solving the equation:

3x - 4 = 8

3x - 4 + 4 = 8 + 4

3x = 12

x = 4

6 0

3 years ago

Read 2 more answers

Jacques and Aaron both started this year at 5.65 feet tall. Jacques has grown 0.2 feet since then. Aaron has grown twice that am

OLga [1]

It is 5.69 positively sure of it good luck on youre quiz if youre taking one!

7 0

3 years ago

the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?

madam [21]

Answer:

About 0.6548 grams will be remaining.

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

$f(t)=a(r)^t$

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

$\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}$

One year has 365 days.

Therefore, the amount remaining after one year will be:

$\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548$

About 0.6548 grams will be remaining.

Alternatively, we can use the standard exponential growth/decay function modeled by:

$f(t)=Ce^{kt}$

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

$255=510e^{38k}$

Solving for k:

$\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)$

Thus, our function is:

$f(t)=510e^{t\ln(.5)/38}$

Then after one year or 365 days, the amount remaining will be about:

$f(365)=510e^{365\ln(.5)/38}\approx 0.6548$

5 0

3 years ago