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3 years ago

Please help i’m struggling so much

Mathematics

2 answers:

Lesechka [4]3 years ago

7 0

3 should be the answer

skelet666 [1.2K]3 years ago

7 0

Answer:

3 will be the answer.

Step-by-step explanation:

Hope it's help you ♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

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MARKING BRAINLIEST Which represents the reflection of f(x) = √x over the x-axis?

Roman55 [17]

Answer:

Table1

-1 undefined since we cannot take the sqrt of a negative number

0 0

1 -1

4 -2

Step-by-step explanation:

f(x) = sqrt(x)

A reflection over the x axis is-f(x)

g(x) = -sqrt(x)

x -sqrt(x)

-1 undefined since we cannot take the sqrt of a negative number

0 0

1 -1

4 -2

4 0

3 years ago

Read 2 more answers

Write a recursive formula for the sequence 15, 26, 48, 92, 180.

12345 [234]

15, 26, 48, 92, 180

  26 - 15 = 13
  48 - 26 = 22
  92 - 48 = 44
180 - 92 = 88

  26 ÷ 15 = 1.73
  48 ÷ 26 = 1.84
  92 ÷ 48 = 1.9167
180 ÷ 92 = 1.95

The recursive formula isn't arithmetic or geometric.

4 0

3 years ago

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Angle α lies in quadrant II , and tanα=−125 . Angle β lies in quadrant IV , and cosβ=35 .

Artist 52 [7]

Answer:

$cos(\alpha+\beta)=\frac{33}{65}$

Step-by-step explanation:

step 1

Find cos α

we know that

$tan^2(\alpha)+1=sec^2(\alpha)$

we have

$tan(\alpha)=-\frac{12}{5}$

substitute

$(-\frac{12}{5})^2+1=sec^2(\alpha)$

$sec^2(\alpha)=\frac{144}{25}+1$

$sec^2(\alpha)=\frac{169}{25}$

$sec(\alpha)=\pm\frac{13}{5}$

Remember that Angle α lies in quadrant II

sec α is negative

$sec(\alpha)=-\frac{13}{5}$

Find the value of cos α

$cos)\alpha)=\frac{1}{sec(\alpha)}$

$cos(\alpha)=-\frac{5}{13}$

step 2

Find sin α

we know that

$tan(\alpha)=\frac{sin(\alpha)}{cos(\alpha)}$

$sin(\alpha)=tan(\alpha)cos(\alpha)$

we have

$tan(\alpha)=-\frac{12}{5}$

$cos(\alpha)=-\frac{5}{13}$

substitute

$sin(\alpha)=(-\frac{12}{5})(-\frac{5}{13})$

$sin(\alpha)=\frac{12}{13}$

step 3

Find sin β

we know that

$sin^2(\beta)+cos^2(\beta)=1$

we have

$cos(\beta)=\frac{3}{5}$

substitute

$sin^2(\beta)+(\frac{3}{5})^2=1$

$sin^2(\beta)=1-(\frac{3}{5})^2$

$sin^2(\beta)=1-\frac{9}{25}$

$sin^2(\beta)=\frac{16}{25}$

$sin(\beta)=\pm\frac{4}{5}$

Remember that

Angle β lies in quadrant IV

sin β is negative

$sin(\beta)=-\frac{4}{5}$

step 4

Find cos(α−β)

we know that

$cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)$

we have

$cos(\alpha)=-\frac{5}{13}$

$cos(\beta)=\frac{3}{5}$

$sin(\alpha)=\frac{12}{13}$

$sin(\beta)=-\frac{4}{5}$

substitute the given values

$cos(\alpha+\beta)=(-\frac{5}{13})(\frac{3}{5})-(\frac{12}{13})(-\frac{4}{5})$

$cos(\alpha+\beta)=(-\frac{15}{65})+(\frac{48}{65})$

$cos(\alpha+\beta)=\frac{33}{65}$

7 0

4 years ago

Help help helphelp help helphelp help helphelp help helphelp help helphelp help help

timama [110]

Answer:

Step-by-step explanation:

20, the coefficient of this expression, is being divided by x, the variable. Therefore, the correct answer is b, division. Hope this helps!

5 0

3 years ago

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X − 5 < 4 OR x − 5 ≥ 7

eduard

Answer:

4x-5=7

Step-by-step explanation:

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3 years ago