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2 years ago

Identify a pair of factors of −35 that has a sum of −2.

Mathematics

1 answer:

Allushta [10]2 years ago

6 0

Answer:

-7 and 5

Step-by-step explanation:

If you multiply -7 and 5, you get -35

Add them, and you get -2.

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3 years ago

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Write a paragraph proof.

worty [1.4K]

It is given that line segment BC is congruent to line segment EC and that line segment AC is congruent to DC. Because of the vertical angles theorem, angle BCA is equal to angle DCE. Therefore, triangles CBA AND DEC are congruent by SAS. Using CPCTC, BA is equal to ED.

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3 years ago

Suppose the average math SAT score for students enrolled at local community college is 490.4 with a standard deviation of 63.7.

lakkis [162]

Answer:

Option B) 9.1

Step-by-step explanation:

We are given the following in the question:

Average score = 490.4

Standard deviation = 63.7

Sample size, n = 49

Formula:

$=\dfrac{\sigma}{\sqrt{n}}$

Putting values, we get,

Standard error =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{63.7}{\sqrt{49}} = 9.1$

Thus, the correct answer is

Option B) 9.1

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3 years ago

Determine the amplitude (A), period (P), and phase shift (PS) of the function f(x)= 3sin(x-4)+9

Tpy6a [65]

ANSWER

D.
$A=3,P=2\pi,PS=(4,9)$

EXPLANATION

The given function is

$f(x) = 3 \sin(x - 4) + 9$

Comparing to the general form of a trigonometric sine function,

$f(x) = A \sin(Bx-C)+D$

$A=3$
The amplitude is absolute value of A

$|A| = |3| = 3$

$B=1$

The period is given by

$p = \frac{2\pi}{ B}$

$p = \frac{2\pi}{ 1} = 2\pi$

$C=4$

is a shift to the right of 4 units.

and

$D=9$

is an upward shift of 9 units

Therefore the phase shift is (4,9).

The correct answer is D

6 0

3 years ago

John ordered a large three topping pizza for $7. 50 and a large 8 topping pizza for $10. Write an equation in slope-intercept fo

yKpoI14uk [10]

to get the equation of any straight line, we simply need two points off of it, well, let's use the provided values hmmm

$\begin{array}{ccll} \stackrel{toppings}{x}&\stackrel{cost}{y}\\ \cline{1-2} 3 & 7.50\\ 8& 10 \end{array} \qquad\implies \qquad (\stackrel{x_1}{3}~,~\stackrel{y_1}{7.50})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{10}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{10}-\stackrel{y1}{7.50}}}{\underset{run} {\underset{x_2}{8}-\underset{x_1}{3}}}\implies \cfrac{2.5}{5}\implies \cfrac{~~ \frac{25}{10}~~}{5}\implies \cfrac{25}{50}\implies \cfrac{1}{2}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{7.50}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{3})\implies y-\cfrac{750}{100}=\cfrac{1}{2}(x-3) \\\\\\ y-\cfrac{15}{2}=\cfrac{1}{2}x-\cfrac{3}{2}\implies y=\cfrac{1}{2}x-\cfrac{3}{2}+\cfrac{15}{2}\implies y=\cfrac{1}{2}x-6$

3 0

2 years ago