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sashaice [31]
3 years ago
6

Plssssss help me plssss

Mathematics
1 answer:
Jlenok [28]3 years ago
5 0

Answer:

here is the answer

Step-by-step explanation:

1. p is the answer

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There are 16 ounces in 1 pound. Juan is mailing a package that weights 10 pounds. He wants to know the weight of the package in
NikAS [45]

Answer:

160 ounces

Step-by-step explanation:

Easy mental math.

16oz × 10 pounds = 160 oz

please learn your school work. I know it's boring, but this is actually stuff you'll use in the real world. this is simple 4th-6th grade work. :)

7 0
3 years ago
ZA and ZB are supplementary angles. If mA = (2x − 9)º and m
Papessa [141]

angle A = 30

(2x-9) = (2x+21)
     +9          +9
 2x=2x+30

-2x  -2x

=30

6 0
1 year ago
HELP PLEASE!!<br> What is the value of x
Licemer1 [7]
X=-8 is your answer for this problem
7 0
3 years ago
Pls, can you help me? thx.<br><img src="https://tex.z-dn.net/?f=%20%5Ccos%28x%20%2B%20%5Cfrac%7B%5Cpi%7D%7B3%7D%29%20%5Cgeqslant
Veseljchak [2.6K]

For x between -\pi and \pi, we have

  • \cos x=\frac{\sqrt2}2=\frac1{\sqrt2} when x=\pm\frac\pi4;
  • \cos x=1 for x=0; and
  • \cos x=0 for x=\pm\frac\pi2

\cos x is continuous over its domain, so the intermediate value theorem tells us that

\cos x\ge\frac{\sqrt2}2

is true for -\frac\pi4\le x\le\frac\pi4.

For all x, we take into account that \cos x is 2\pi-periodic, so the above inequality can be expanded to

-\dfrac\pi4\le x+2n\pi\le\dfrac\pi4

where n is any integer. Equivalently,

-\dfrac\pi4-2n\pi\le x\le\dfrac\pi4-2n\pi

To get the corresponding solution set for

\cos\left(x+\dfrac\pi3\right)\ge\dfrac{\sqrt2}2

simply replace x with x+\frac\pi3:

-\dfrac\pi4-2n\pi\le x+\dfrac\pi3\le\dfrac\pi4-2n\pi

\implies\boxed{-\dfrac{7\pi}{12}-2n\pi\le x\le-\dfrac\pi{12}-2n\pi}

7 0
3 years ago
Find the zeros of f(x) = -x^2 + 4x - 3 by using a table and a graph.
ludmilkaskok [199]
1) <span>Enter the missing numeral of the answer 1, 3. Use a number only.
2) </span>f(x) = x^2 - 5x - 6 can be written as <span>f(x) = x^2 - 5x - 6 = (x-1) (x+6)
the zeros are 1 and 6, and the smallest zero is 1,
3) </span>g(x) = x^2 - 8x =<span>g(x) = x(x - 8), so the zeros are 0 and 8, the smallest zero is 0
4) </span>f(x) = x^2 - 7x + 6 can be written as <span>f(x) = x^2 - 7x + 6 = (x-1) (x-6), the zeros are 1 and -6, and the smallest zero is 1
5) </span>g(x) = x^2- 6x - 16 can be written as <span>g(x) = x^2- 6x - 16 = (x+2) (x-8)
the zeros are -2 and 8, the smallest zero is -2</span>
7 0
3 years ago
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