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3 years ago

Estimate the solution to the system of equations.

Mathematics

1 answer:

Natali [406]3 years ago

3 0

Answer: y=14/3, x= 5/3

Step-by-step explanation:

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What does x equal in x+5=-10

Zigmanuir [339]

Answer:

-15

Step-by-step explanation:

-15+5=-10

5 0

4 years ago

Read 2 more answers

Please answer these for me

raketka [301]

Well I know that the square root of 121 is 11

7 0

4 years ago

Read 2 more answers

Find the product. (-5a2)3 · a5

lesya692 [45]

Answer:

$-125a^{11}$

Step-by-step explanation:

The expression that we have in this problem is

$(-5a^2)^3\cdot a^5$

First of all, we proceed by evaluating the power of the term in the brackets. We apply the rule:

$(a^m)^n=a^{m\cdot n}$

$(a^2)^3=a^6$

Moreover, we have

$(-5)^3=-125$

Therefore,

$(-5a^2)^3=-125 a^6$

So now the expression is

$-125a^6 \cdot a^5$

Now we apply the following rule:

$a^m \cdot a^n = a^{m+n}$

So we have

$a^6\cdot a^5 = a^{6+5}=a^{11}$

Therefore, our expression becomes:

$-125a^6\cdot a^5=-125a^{11}$

3 0

4 years ago

What is the value of es026-1.jpg?

Kay [80]

Answer:

Value of $[(\frac{2}{3}) ^0]^{-3}$ is, 1

Step-by-step explanation:

To find the value of $[(\frac{2}{3})^0]^{-3}$

Using exponents power rule:

$(a^m)^n = a^{mn}$

then;

$(\frac{2}{3})^{0 \cdot (-3)}$

Simplify:

$(\frac{2}{3})^{0}$

By Zero exponent rule: Anything raised to the power 0 is 1. i.e,

$a^0 = 1$

then;

$(\frac{2}{3})^{0}$ = 1

Therefore, the value of $[(\frac{2}{3}) ^0]^{-3}$ is, 1

4 0

3 years ago

Express the area A of a circle as a function of its radius, r. Determine the area of a circle whose radius is 11 inches. That

nekit [7.7K]

<h2>Answer with explanation:</h2>

The function represents the area of a circle(A) is given by :-

$A(r)=2\pi r$

To determine the area of a circle whose radius is 11 inches. , we put the value of r=11 inches, we get

$A(11)=2\pi (11)$

Now, we put $\pi=\dfrac{22}{7}$ in the above function , we get

$A(11)=2(\dfrac{22}{7} )(11)\\\\\Rightarrow\ A=56.5714285714\approx56.57$

8 0

3 years ago