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2 years ago

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1. Solve the problem [(–7 + 2) + 4] – 3 = ?

1 answer:

Anna11 [10]2 years ago

4 0

For number 3 it is D

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ra1l [238]

<h2>Explanation:</h2>

Let's use a trial and improvement method to find this solution.

Step 1. Let's choose x = 8.5

Substituting into the equation:

$(8.5)^2+ 3(8.5) - 94=0? \\ \\ 72.25+25.5-94=0? \\ \\ 3.75\neq 0$

Step 2. Let's choose x = 8.4

Substituting into the equation:

$(8.4)^2+ 3(8.4) - 94=0? \\ \\ 70.56+25.2-94=0? \\ \\ 1.76\neq 0$

Step 3. Let's choose x = 8.3

Substituting into the equation:

$(8.3)^2+ 3(8.3) - 94=0? \\ \\ 68.89+24.9-94=0? \\ \\ -0.21\neq 0$

Since the sign of the equation changes from positive to negative when evaluating from 8.4 to 8.3, then x = 8.3 seems to be a reasonable value. Finally, the solution to 1 decimal place is:

$x=8.3$

6 0

2 years ago

ONE more question Ah

Setler [38]

Answer: I don’t understand?

Step-by-step explanation:

6 0

2 years ago

On Saturday morning, Malik earned $32. By the end of the afternoon he had earned a total of $63. Which equation represents this

aleksley [76]

X+ 32= 63 that is correct answer

6 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

How many lines of symmetry does the figure have?<br><br> A. 0<br> B. 1<br> C. 2

balu736 [363]

Answer:

The answer is A, so zero.

8 0

2 years ago