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2 years ago

6

2x-16 in words phases

1 answer:

Marrrta [24]2 years ago

4 0

Answer:

16 less than two times a number "x"

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What is 4/5 of $45.00

Alex73 [517]

I did this in my head but the answer is $36 45/5 is 9 9*4 is 36. And if you add another it totals to 45. Thus the answer is $36.

4 0

3 years ago

The ratio of male students to female students in the drama club at Campo high school is 3:4. If the number of male students in t

alexgriva [62]

Set up an equation. Since the ratio is 3:4 male:female, it would be 3m=4f. Since you know that 3m=18, solve for m and plug it into the 3m=4f equation to get the number of females! Don't forget that the number of females is 4f not just f!

3 0

4 years ago

Solve the triangle. Round your answers to the nearest tenth.

mixas84 [53]

Answer:

Angles: $\angle A = 43^{\circ}$ $\angle B = 55^{\circ}$ $\angle C = 82^{\circ}$

Sides: $BC = 20$ $AB = 29$ $AC =24$

Step-by-step explanation:

See attachment for complete question

From the attachment, we have:

$AB = 29$

$AC =24$

$\angle C = 82^{\circ}$

Required

Complete the missing side and missing angles

To calculate angle B, we apply sine laws:

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

In this case:

$\frac{AB}{sinC}=\frac{AC}{sinB}$

This gives:

$\frac{29}{sin(82^{\circ})}=\frac{24}{sinB}$

$\frac{29}{0.9903}=\frac{24}{sinB}$

Cross Multiply

$sinB * 29 = 24 * 0.9903$

Divide both sides by 29

$sinB = \frac{24 * 0.9903}{29}$

$sinB = 0.8196$

Take arcsin of both sides

$B = sin^{-1}(0.8196)$

$B = 55^{\circ}$

So:

$\angle B = 55^{\circ}$

To solve for the third angle, we make use of:

$\angle A + \angle B + \angle C = 180^{\circ}$

This gives:

$\angle A + 55^{\circ} + 82^{\circ} = 180^{\circ}$

$\angle A + 137^{\circ}= 180^{\circ}$

$\angle A = 180^{\circ}- 137^{\circ}$

$\angle A = 43^{\circ}$

Hence, the angles are:

$\angle A = 43^{\circ}$ $\angle B = 55^{\circ}$ $\angle C = 82^{\circ}$

To calculate the length of the third side, we apply cosine law

$BC^2 = AB^2 + AC^2 - 2*AB*AC*cosA$

$BC^2 = 29^2 + 24^2 - 2*29*24*cos(43^{\circ})$

$BC^2 = 841+ 576 - 1392*cos(43^{\circ})$

$BC^2 = 841+ 576 - 1392*0.7314$

$BC^2 = 841+ 576 - 1018.11$

$BC^2 = 398.89$

Take the square root of both sides

$BC = \sqrt{398.89$

$BC = 19.9722307217$

$BC = 20$

5 0

3 years ago

Please can you answer It is due very soooooonn Will mark brainiest

Assoli18 [71]

Answer:

Mean of a grouped data is

Sum of F x/ sum of frequency F

Sum of F = 5 + 15 + 13 + 10 + 7 = 50

to find x find the average of the two classes

That's

0 + 10/2 = 10/2 = 5

10+20/2 = 15

20 + 30 / 2 = 25

30+40 /2 = 35

40+ 50 / 2 = 45

Therefore sum of Fx = 5(5) + 15(15) + 13(25) + 10(35) + 7(45)

= 1240

Therefore

Mean = 1240/50

= 24.8cm

I hope this helps you

4 0

3 years ago

What value of y makes the equation true? -7/12=y/-36

STALIN [3.7K]

-7/12=y/-36
times both sides by -36 (-36=12*-3)
-3*-7=y
21=y

y=-21

7 0

3 years ago