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2 years ago

The expression 9x ^(2)-5x -7 has_ terms and a constant of __ .

Mathematics

1 answer:

Elenna [48]2 years ago

8 0

The expression $9x ^2-5x -7$ has_3__ terms and a constant of __-7__ .

Step-by-step explanation:

The expression $9x ^2-5x -7$ has___ terms and a constant of ____ .

Each term in an expression is separated by + or -

A constant is a term in an expression that contains only numbers. It doesn't have variables in it.

So, in the given expression $9x ^2-5x -7$

Terms : 3 i.e, 9x^2, 5x, 7

and constant = -7

The expression $9x ^2-5x -7$ has_3__ terms and a constant of __-7__ .

Keywords: Terms and Constants

Learn more about Terms and Constants at:

#learnwithBrainly

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2 years ago

What are the y-intercept of the function f(x)=-2x^2-3x+20?

Serhud [2]

Answer:

$\sf A) \ (-4, \ 0) \ and \ \ [\dfrac{5}{2} , \ 0]$

Explanation:

Given function: f(x) = -2x² - 3x + 20

To find the x-intercepts of a function, f(x) = 0

=================

-2x² - 3x + 20 = f(x)

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2 years ago

Mr. Burn's class is taking a field trip to the planetarium. The trip will cost $27 for each

marysya [2.9K]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

likoan [24]

It's easy to show that $7\tan(4x)$ is strictly increasing on $x\in\left[0,\frac\pi8\right]$ . This means

$M = \max \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/12} = 7\sqrt3$

and

$m = \min \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/16} = 7$

Then the integral is bounded by

$\displaystyle 7\left(\frac\pi{12} - \frac\pi{16}\right) \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le 7\sqrt3 \left(\frac\pi{12} - \frac\pi{16}\right)$

$\implies \displaystyle \boxed{\frac{7\pi}{48}} \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le \boxed{\frac{7\sqrt3\,\pi}{48}}$

7 0

2 years ago

The expression 9x ^(2)-5x -7 has___ terms and a constant of ____ .

The expression 9x ^(2)-5x -7 has_ terms and a constant of __ .