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2 years ago

8

Pls answer this i will mark brainlest

1 answer:

vodomira [7]2 years ago

7 0

Answer:

Answer is a

Step-by-step explanation:

it is the only answer that makes sense

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G<br> Express as a percent.<br> 4/5%

Aloiza [94]

Answer:

80%

Step-by-step explanation:

I'm assuming you meant express 4/5 as a percent and not 4/5%. So simply evaluate 4/5 to get 0.80. Now multiply this value by 100 to get the percentage which gives you 80%

5 0

2 years ago

5. photo album: $25.50, 10% markup

sasho [114]

Answer:

$28.05

Step-by-step explanation:

If you want to find how much the photo album is after the markup, you first have to find 10% of the original price. 10% of $25.50 is $2.55, so you add $2.55 and $25.50 to find the price after the markup.

Hope this helps!

3 0

3 years ago

Solve for x ‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎ ‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎ ‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎ ‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎

inysia [295]

Answer:

12.699

Step-by-step explanation:

i think if you do 26-22.4, you get the leg of the small triangle which is 3.6

then do Pythagorean theorem

3.6^2 + x^2 = 13.2^2

12.96 +x^2 = 174.24

174.24-12.96

x^2 = 161.28

take square root

x = 12.699

6 0

2 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

Need help As soon as possible!!

emmasim [6.3K]

I think the fourth option is the answer. It wi give you the Side angle side postulate.

4 0

3 years ago