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3 years ago

Determine the number of integers between -10 and 10.

Mathematics

1 answer:

RideAnS [48]3 years ago

8 0

There are 20 integers between -10 and 10

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Find the product. -8x5y2 · 6x2y

katrin [286]

Answer:

The product of the given expression $-8x^5y^2\times 6x^ 2y$ = 72x^6y^3[/tex]

Step-by-step explanation:

Given expression is $-8x^5y^2\times 6x^2y$

To find the product of the given expression:

The product of the given expression can be written as

$-8x^5 y^2\times 6x 2y=(-8x^5y^2)(6x^2y)$

$=(-8x^5y^2)(6x^2y)$

$=-48(x^5\times x^2)(y^2\times y^1)$ [since $a^m\times a^n=a^{m+n}$ ]

$=-48x^7y^3$

Therefore the product of the given expression $-8x^5y^2\times 6x^2y$ is $-48x^7y^3$

Therefore, $-8x^5y^2\times 6x^2y$ =-48x^7y^3[/tex]

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3 years ago

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zvonat [6]

Answer:7.375?

Step-by-step explanation:

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3 years ago

If two triangles are congruent, then the six ___ parts are congruent.

mel-nik [20]

If two triangles are congruent, then the six corresponding parts are congruent.

5 0

3 years ago

Read 2 more answers

What is the LCD of 8 5/12 and 3 5/8? A.24 B.96 D.48 C.16

Rama09 [41]

Answer:

Step-by-step explanation:

12, 24

8, 16, 24

6 0

3 years ago

Find K if x + 1 is a factor of P(x) = Kx^2-x+2

wlad13 [49]

Answer:

$K=-3$

Step-by-step explanation:

According to the Factor Theorem, a binomial (x - a) is a factor of a polynomial P(x) if and only if P(a) = 0.

We have the polynomial:

$P(x)=Kx^2-x+2$

And we want to find the value of K such that (x + 1) is a factor.

We can rewrite our factor as (x - (-1)). Thus, a = -1.

Then in orer for (x + 1) to be a factor, P(-1) must equal 0. Hence:

$P(-1)=K(-1)^2-(-1)+2=0$

Simplify:

$K+1+2=0$

Solve for K. Hence:

$K=-3$

8 0

3 years ago