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2 years ago

Write the espression 40w - 16, as a product with a whole number greater then one.

Mathematics

2 answers:

musickatia [10]2 years ago

8 0

Answer:

8(5w - 2).

Step-by-step explanation:

40w - 16

The greatest common factor of the two terms is 8 so the product is:

8(5w - 2).

pashok25 [27]2 years ago

6 0

Answer:

8(5w-2)

Step-by-step explanation:

40w-16

Since the GCF is 8, hence we take 8 common

8(5w-2)

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Heather made a total of $451.78 last month at her part-time job. If she worked 14 hours, how much did she make per hour?

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The amount of money divided by the amount of hours will give you how much money she makes per hour.

451.78/14=32.27.

She earned $32.27 per hour.

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2 years ago

Which expression is equivalent to .........? <br> ......<br> ......<br> .....<br> .

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Answer:

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Step-by-step explanation:

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2 years ago

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Linda bought a cube jewelry box. The volume of the jewelry box is 64 cubic inches. What is the edge length of Linda’s jewelry bo

yawa3891 [41]

Answer:

The edge length will be equal to 4 inches.

Step-by-step explanation:

The volume of a cube = s * s * s (in which s represent side of the cube)

V = s * s * s

V = s³

64 = s³

64 = 4³

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2 years ago

A store purchased a picnic basket for $9.88 and marked it up 55%. During a sale, the store marked it down 5%. What was the disco

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14.82

Step-by-step explanation:

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2 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0

2 years ago