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3 years ago

The lowest point in a business cycle, which follows a period of economic

Mathematics

2 answers:

Anika [276]3 years ago

7 0

Your answer is b hope this helps

Nonamiya [84]3 years ago

4 0

Answer: it’s a trough

Step-by-step explanation:

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Max claims that a point on any line that is perpendicular to a segment is equidistant from a segment's endpoints. Charlene claim

yan [13]

Answer:

Charlene claim is true.

Step-by-step explanation:

Max claims that a point on any line that is perpendicular to a segment is equidistant from a segment's endpoints.

It is not necessary as shown in the diagram (a).

Charlene claims that the line must be a perpendicular bisector for a point on the line to be equidistant from a segment's endpoints.

It is true as shown in the diagram (b).

So, Charlene claim is true.

3 0

3 years ago

Select all the numbers that belong to the special sequence of square numbers.

ipn [44]

answer:

169

81 =9²

169=13²

7 0

3 years ago

Write the linear function for the given function values.<br> f(-2)=9 and f(6)=7

BigorU [14]

Answer:

$y = -\frac14x + 8.5$

Step-by-step explanation:

Hello!

A linear function is written in the form of $y = mx + b$ , where m is the slope and b is the y-intercept.

Each of these expressions can be converted to coordinate form, (-2,9) and (6,7).

<h2>Slope</h2>

Slope can be calculated by the difference in the x-values divided by the difference in the y-values.

Solve

$\frac{9 - 6}{-2 - 6}$
$\frac{2}{-8}$
$-\frac14$

The slope is -1/4.

<h2 /><h2>Y-intercept</h2>

The y-intercept is found by plugging in the x and y values of a coordinate and the slope and then solving for b.

Solve

$y = mx + b$
$9 = -\frac14(-2) + b$
$9 = 0.5 + b$
$8.5 = b$

<h2>Equation</h2>

We can find the equation by simply plugging in all the values that we found.

Equation: $y = -\frac14x + 8.5$

6 0

2 years ago

-3 is less than or equal to W and W is less than or equal to 4

ANEK [815]

Answer:

-3 ≤ W ≤ 4

I hope this helps!

8 0

4 years ago

Expand using the properties and rules for logarithms

malfutka [58]

Consider expression $\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).$

1. Use property

$\log_a\dfrac{b}{c}=\log_ab-\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.$

2. Use property

$\log_abc=\log_ab+\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.$

3. Use property

$\log_ab^k=k\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.$

4. Use property

$\log_{a^k}b=\dfrac{1}{k}\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.$

Answer: correct option is B.

7 0

4 years ago