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4 years ago

12

How many yards a football team moves on a field using this expression |8| + |-4|

1 answer:

Harman [31]4 years ago

7 0

Answer: its 12 because even though it says -4 the lines around it just make it 4 so 8+4=12

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What are the solutions to the quadratic equation x^2-16=0

frutty [35]

Simplifying this, we get:
(x+4)(x-4) = 0
Now, to get 0, we can replace x for one of these things to make 0, so:
x can either be 4, or -4

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3 years ago

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7a^2/9b^4 simplify the fraction

CaHeK987 [17]

$7* \dfrac{a^2}{9}b^4$

$\dfrac{7a^2b^4}{9}$

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3 years ago

Select the correct answer from the drop-down menu.

ELEN [110]

Missing term = –2xy

Solution:

Let us first find the quotient of $-8x^2y^3 \div xy$ .

$-8x^2y^3 \div xy=\frac{-8x^2y^3 }{xy}$

$=\frac{-8\times x\times x\times y\times y\times y}{xy}$

Taking common term xy outside in the numerator.

$=\frac{xy(-8\times x\times y\times y)}{xy}$

Both xy in the numerator and denominator are cancelled.

$=-8xy^2$

Thus, the quotient of $-8x^2y^3 \div xy$ is $-8xy^2$ .

Given the quotient of $-8x^2y^3 \div xy$ is same as the product of 4xy and ____.

$-8xy^2=4xy$ × missing term

Divide both sides by 4xy, we get

⇒ missing term = $\frac{-8xy^2}{4xy}$

Cancel the common terms in both numerator and denominator.

⇒ missing term = –2xy

Hence the missing term of the product is –2xy.

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3 years ago

Which graph does NOT represent a function?

sergeinik [125]

Answer:

the second one

i just did it

Step-by-step explanation:

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Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2

NNADVOKAT [17]

$y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}$

Divide both sides by $\dfrac13y^{-2}(x)$ :

$3y^2y'-y^3=xe^x\ln x$

Substitute $v(x)=y(x)^3$ , so that $v'(x)=3y(x)^2y'(x)$ .

$v'-v=xe^x\ln x$

Multiply both sides by $e^{-x}$ :

$e^{-x}v'-e^{-x}v=x\ln x$

The left side can be condensed into the derivative of a product.

$(e^{-x}v)'=x\ln x$

Integrate both sides to get

$e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C$

Solve for $v(x)$ :

$v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

Solve for $y(x)$ :

$y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

$\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}$

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3 years ago