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2 years ago

Please help fast !!!! Five times as much as the quotient of 96 and 3 what is the answer?

Mathematics

2 answers:

inessss [21]2 years ago

8 0

Answer:

1440

Step-by-step explanation:

slamgirl [31]2 years ago

3 0

Answer:

hiiii

Step-by-step explanation:

1440

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What is the slope of the line that passes through the points (1,7) and (-4,-8)?

Aleksandr [31]

Answer:

D. 3

Step-by-step explanation:

To find the slope of a line, we can use the equation y2-y1/x2-x1=m or slope.

All we can do now is insert the values of the points into the equation.

-8-7/-4-1=m

-15/-5=m

3=m

Thus, the slope is 3.

Hope this helps!

5 0

3 years ago

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Find the measure of the missing angle.

ICE Princess25 [194]

Answer:

36 degrees

Step-by-step explanation:

The whole shape is 90 degrees.

90 - 54 = 36

36 degrees

6 0

3 years ago

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I have done 14 projects so far, and gotten 100% on all of them.

mariarad [96]

100x14=1400
1400 divided by 15=93.3
So your grade would still be a low A

5 0

2 years ago

1. Denise makes 21 quarts of ice cream. She stores the ice cream in containers

Mkey [24]

Answer:

21 containers

Step-by-step explanation:

21/1=1

4 0

3 years ago

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

8 0

3 years ago

Read 2 more answers