Ask question

Ask question

All categories

IgorLugansk [536]

3 years ago

14

Solve this equation 5x-7=-12

2 answers:

cupoosta [38]3 years ago

7 0

Answer:

5x - 7= -12

5x - 7+7 = -12+ 7

5x= -5

x= -1

Step-by-step explanation:

hope this helped

Tasya [4]3 years ago

4 0

Answer:

<u><em>x = -1</em></u>

Step-by-step explanation:

Isolate the variable by adding 7 on both sides. 5x-7+7 is 5x, and -12+7 is -5. Then, you get the equation 5x = -5. You have to then divide by 5 on both sides. 5x/5 is x and -5/5 is -1. You would then get that x = -1.

You might be interested in

Choose the answers that have a repeating decimal?

Dmitry [639]

A 0.66... that the repeating decimals

4 0

3 years ago

45.Erica Glaser deposits a check for $20,005.00 and $1.234.00 in cash. What is het total

Gwar [14]

Her total is $21,239

5 0

2 years ago

Help me i mark brain thing help due in 5 minutes

Dmitry_Shevchenko [17]

Answer:

A: 1

B: 17

C: 26

D: 42

E: 57

Step-by-step explanation:

4 0

3 years ago

Two and one third minus two thirds

Savatey [412]

The answer to 2 1/3 - 2/3 is 1 2/3

4 0

1 year 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