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3 years ago

10

Need help pls 8th math

1 answer:

finlep [7]3 years ago

3 0

I am not completely sure but it's most likely the last option because congruent means equal and the two trapezoids can't be equal because of the different side lengths.

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-2/3 - (-1 1/3)<br><br> A -1 2/3<br><br> B -2/3<br><br> C-1/3<br><br> D 2/3

mihalych1998 [28]

The answer to your question is D , it is 2/3 simple math even easier if you use a calcul

5 0

3 years ago

Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

Answer:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}} = 32$

Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}}$

We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

3 0

3 years ago

Warm up question, what is 5/6 + 5/6? Explain how you got the answer.

Helga [31]

Answer: 5/3

Step-by-step explanation:

6 0

3 years ago

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What is the simplified fraction for 6%?

mr Goodwill [35]

Answer:

6/10?

Step-by-step explanation:

6 0

3 years ago

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Write the equation of a circle with a center at (-1,2) and a radius of 7

Sever21 [200]

Answer:

$(x+1)^2+(y-2)^2=49$

Step-by-step explanation:

The standard form for the equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where r is the radius, and the center is (h, k).

(h, k) = (-1, 2), r = 7

$(x-(-1))^2+(y-2)^2=7^2\\\\(x+1)^2+(y-2)^2=49$

5 0

3 years ago