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3 years ago

Write the sum as a product of the GCF of 96 + 18?

Mathematics

1 answer:

worty [1.4K]3 years ago

7 0

96: 1,2,3,4,6,8,12,16,24,32,48,96

18: 1,2,3,6,9,18

GCF: 6

But, if you're asking what 96 + 18 it's 114

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babunello [35]

Answer: I do not understand the options you gave but but cos30º is equal to $\sqrt{3}$ /2 which is 0.86602540378.

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3 years ago

How many solutions does the equation have? 0 = -8q + 8q

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3 years ago

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F(n) = n² – 3 g(n) = 4n - 1 Find f[g(1)]

SVETLANKA909090 [29]

Answer:

f[g(1)]=6.

Explanation:

Given f(n) and g(n) defined below:

$\begin{gathered} f\mleft(n\mright)=n^2-3 \\ g\mleft(n\mright)=4n-1 \end{gathered}$

First, we evaluate g(1):

$\begin{gathered} g\mleft(1\mright)=4(1)-1 \\ =4-1 \\ g(1)=3 \end{gathered}$

Therefore:

$\begin{gathered} f\mleft(g(1)\mright)=f\mleft(3\mright) \\ f\mleft(3\mright)=3^2-3 \\ =9-3 \\ =6 \end{gathered}$

Therefore, f[g(1)]=6.

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1 year ago

What equations have z=4

natita [175]

<h2>Just some examples...</h2>

2z+4=12

2(4)+4=12

____________

5z*2=40

5(4)*2=40

<h2 />

6 0

3 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

3 years ago