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Vladimir79 [104]

2 years ago

6

What is 5÷1000 rounded to the nearest hundredth i need help ASAP

2 answers:

REY [17]2 years ago

7 0

Answer:

0.005

Step-by-step explanation:

Marat540 [252]2 years ago

3 0

0.005. If i'm wrong, can ou tell me the answer choices?

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What is the answer to the math problem 18 minus 4x = negative 2?

pentagon [3]

The answer to this equation is 5.

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3 years ago

Use the area of a rectangle and the Distributive<br> Property to find each product.<br> 9 X 34

OLga [1]

Answer:

Step-by-step explanation:

Write 34 in tens and units.

34 = 30 + 4

9(30 + 4)

9 * 30 = 270

9* 4 = 36

270 + 36 = 306

After awhile if your practice enough, you can do them in your head.

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2 years ago

A pizza restaurant sells containers of juice that it marks up 75 percent above its wholesale cost. If the restaurant marks up ea

Len [333]

75% of what is 0.90
0.75x = 0.90
x = 0.90/0.75
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4 0

3 years ago

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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

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the shaded area of the following shape can be found by subtracting the area of the circle from the area of the triangle?

balu736 [363]

I found the accompanying picture of this word problem.

The circle was inscribe in the triangle. The area outside of the circle but inside the triangle is the shaded area.

True. The shaded area of the following shape can be found by subtracting the area of the circle from the area of the triangle.

7 0

3 years ago