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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! Divide.

Mathematics

1 answer:

ikadub [295]3 years ago

3 0

Answer:

The correct answer is marked.

Step-by-step explanation:

The solution is easily found using synthetic division. The attachment shows how that is done, and what the result is.

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If the slope of a line is -1/4,<br>find x when (x,2) and (1/2,6)

Alecsey [184]

Use slope formula:
m = (y2-y1) / (x2-x1)
-1/4 = 6-2 / 1/2 - x
-1/4 = 4 / 1/2 - x
-1(1/2 - x) = 4(4)
-1/2 + x = 16

x = 16 1/2 or 33/2

7 0

3 years ago

According to the manufacturer of the candy Skittles, 20% of the candy produced are red. If we take a random sample of 100 bags o

igomit [66]

Answer:

Probability that the proportion in our sample of red candies will be less than 20% is 0.5 .

Step-by-step explanation:

We are given that 20% of the candy produced are red. A random sample of 100 bags of Skittles is taken.

The distribution we can use here is;

$\frac{\hat p - p}{\sqrt{\frac{\hat p (1- \hat p)}{n} } }$ ~ N(0,1)

where, p = 0.20 and n = 100

Let $\hat p$ = proportion of red candies in our sample

So, P( $\hat p$ < 0.20) = P( $\frac{\hat p - p}{\sqrt{\frac{\hat p (1- \hat p)}{n} } }$ < $\frac{0.20 - 0.20}{\sqrt{\frac{0.2 (1- 0.2)}{100} } }$ ) = P(Z < 0) = 0.5

Therefore, probability that the proportion in our sample of red candies will be less than 20% is 0.5 .

5 0

3 years ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago

Simplify the following functions so that your answers do not include trig functions. In each case, assume x∈(−1,1). cos(2sin−1x−

Roman55 [17]

Answer:

cos(2sin−1x)=1−2x2

Step-by-step explanation:

et sin−1x=θ , so that, x=sinθ,|x|≤1,θ∈[−π2,π2].

3 0

3 years ago

Solve 25n=2×25 for n. You can use a related equation. n = <img src="https://tex.z-dn.net/?f=%5Cfrac%7B2%2A25%7D%7B25%7D" id="Tex

Archy [21]

Answer:

n=2

Step-by-step explanation:

25n=2×25

Divide each side by 25

25n/25=2×25/25

n = 2 * 25/25

n =2

3 0

3 years ago