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2 years ago

Thrice the difference between x and 4 is less than9 what's the inequality

1 answer:

4 0

(x-4)•2 < 9
the difference between x and 4 means to subtract
and twice that means to double or times by 2
and all that’s LESS than 9
so the sign opens up towards the 9

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Plz help I need bt tonight

JulijaS [17]

The answer is 25%.

First: work out the difference between the two numbers you are comparing.

Decrease = Original Number - New Number

35-28=7

Then: divide the decrease by the original number and multiply the answer by 100. % Decrease = Decrease ÷ Original Number × 100

7÷

28=.25

.25*100=25

<h2>Answer is 25%.</h2>

7 0

2 years ago

Is 3/3 a proper fraction

nexus9112 [7]

No I hope that helps

3 0

2 years ago

Can someone please help!

vekshin1

Answer:

Step-by-step explanation:

Area of a Triangle = Base * Height * 1/2

Base = 5

Height = 4.33

Area of a Triangle = 5 * 4.33 * 1/2

Area of a Triangle = 10.825

There are four triangles.

Surface Area = Area of a Triangle * Number of Triangles

Surface Area = 10.825 * 4

Surface Area = 43.3cm²

7 0

2 years ago

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>

asambeis [7]

Given:

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

To prove:

The given statement.

Proof:

We have,

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

$LHS=\cos^4 \alpha+\sin^4\alpha$

$LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2$

$LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha$ $[\because a^2+b^2=(a+b)^2-2ab]$

$LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha$ $[\because \cos^2 \alpha+\sin^2\alpha=1]$

$LHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

Now,

$RHS=\dfrac{1}{4}(3+\cos 4 \alpha)$

$RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]$

$RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]$ $[\because (a-b)^2=a^2-2ab+b^2]$

$RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]$

$RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]$

$RHS=1+2\cos^4 \alpha-2\cos \alpha$

$RHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

$LHS=RHS$

Hence proved.

8 0

2 years ago

Number 17 pls help me it’s my last question plss

Sonbull [250]

Answer:

Step-by-step explanation:

7 0

2 years ago