Halp please and show you work cuz I'm nervous TwT

never [62]

0.3 or 30% is the same as 3/10

7 0

2 years ago

show work please show work please show work please show work please show work please show work please show work please show work

tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

35% means 35 per hundred. 35/100. simplify and write as 7/20.

0.64 means its 2 zeros left of regular number. 64/100.

82% means 82 per hundred. 82/100. move decimal 2 spaces to left (division is left, multiplication is right). 0.82.

4/5. try to make denominator 100. multiply both numerator and denominator by 20. 80/100. move decimal 2 spaces to the left. 0.8.

14/50. try to make denominator 100. multiply both numerator and denominator by 2. 28/100. 28%

0.19 can be written as 19/100. 19%

7 0

3 years ago

Read 2 more answers

Write the slope intercept form of an equation with a slope of 5 and y-intercept of -4

Delvig [45]

Answer:

y = 5x - 4

Step-by-step explanation:

5 is the slope so it goes in front of the x and -4 is the y Intercept so it goes last.

5 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

2 years ago

Read 2 more answers

Need help with option 1 or 2

Pani-rosa [81]

Option 2.

1. Simplify 7 + (12 - 9) + 7 × 3(8 - 5)

According to the Order of Operations, contents of parentheses are evaluated first. This gives

... 7 + 3 + 7 × 3 × 3

Then multiplication and division are performed left to right.

... = 7 + 3 + 21 × 3

... = 7 + 3 + 63

Followed by addition and subtraction left to right.

... = 10 + 63

... = 73

A Google search box can be relied upon to do the operations in the correct order.

_____

2. When the expression is rewritten, a different result is obtained. This is because the operations indicated by the second set of parentheses are altered.

... (7 + 12) - 9 + (7 × 3)8 - 5

... = 19 - 9 + 21 × 8 - 5

... = 19 - 9 + 168 - 5

... = 10 + 168 - 5

... = 178 - 5

... = 173

In the first expression, both +8 and -5 are multiplied by 21. In the rewritten expression, only +8 is multiplied by 21.

7 0

3 years ago