All categories

3 years ago

A=2 b=3 and c= -1 then find 2^a + 3^b - 5^c

1 answer:

4 0

Just substitute the numbers into its respective letters and do the math.

2^(2) + 3^(3) - 5^(-1)

=4 + 9 - 0.2
=13 + 0.2
=13.2

You might be interested in

If T(x, y) = (x + 5, y + 6) and Pris the image of P, what is the rule for the translation in which P is the image of P'? T(x, y)

xxTIMURxx [149]

Answer:

The translation is;

$T(x,y)=(x-5,y-6)$

Explanation:

Given the translation rule;

$T(x,y)=(x+5,y+6)$

when P' is the image of P.

For the inverse, when P is the image of P', the Translation rule would become;

$\begin{gathered} x=x^{\prime}+5 \\ x^{\prime}=x-5 \\ y=y^{\prime}+6 \\ y^{\prime}=y-6 \\ So,\text{ the translation rule becomes;} \\ T(x,y)=(x-5,y-6) \end{gathered}$

The translation is;

$T(x,y)=(x-5,y-6)$

3 0

1 year ago

Which is an equation in slope intercept form of the line that contains the points (7,-9) and (8,-7)

jeyben [28]

Answer:

y=2x-23

Step-by-step explanation:

m=(-9+7)/(7-8)=-2/-1=2

y=2x+b

-7=2(8)+b

-7-16=b

b=-23

y=2x-23

7 0

3 years ago

Total pay before deductions is known as . net pay gross pay total pay

Effectus [21]

Gross pay.................

6 0

3 years ago

Read 2 more answers

The number math problems Michel can solve varies directly with the amount of time he spends solving math problems. Michel can so

Mama L [17]

Answer:

40 problems

Step-by-step explanation:

8 problems : 5 mins

? : 25 mins

8 x 5 = 40

40 problems

hope this helps

braonliest plz?

5 0

3 years ago

Read 2 more answers

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

7 0

3 years ago

Read 2 more answers