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3 years ago

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I need help with this math problem! Thanks! -8 = -( x +4)

1 answer:

Damm [24]3 years ago

7 0

Answer:

x = 4

Step-by-step explanation:

If the lonely negative sign infront of the brackets is confusing, it can be rewritten as -1.

To solve, simplify by expanding then isolate x.

-8 = -( x +4)

-8 = -1 ( x +4) Distribute over brackets by changing the each term's sign

-8 = -x - 4 Start isolating x by moving everything to the other side

-8 +4 = -x -4 + 4 Add 4 to both sides so it cancels out on the right

-4 = -x

-4 = -1x <= -x is the same at -1x

-4/-1 = -x/-1 <=divide both sides by negative 1

4 = x

x = 4 <= usually we write the variable on the left side

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Solve each compound inequality. Graph your solutions.<br> 5 + m > 4 or 7m< -35

Yuliya22 [10]

Answer:

Solving the inequality $5 + m > 4\: or\: 7m< -35$ we get $\mathbf{m>-1\:or\: \:m$

The graph of the inequality is shown in figure attached.

Step-by-step explanation:

We need to Solve each compound inequality $5 + m > 4\: or\: 7m< -35$

Solving the inequalities

For solving compound inequalities we solve both inequalities simultaneously and find the value of m.

For solving 5+m > 4, we subtract 5 from both sides.

While solving 7m < -35, we divide both sides by 7 to get value of m

Applying these functions we get:

$5 + m > 4\: or\: 7m< -35\\m>4-5\:or\:m-1\:or\:m$

So, solving the inequality $5 + m > 4\: or\: 7m< -35$ we get $\mathbf{m>-1\:or\: \:m$

The graph of the inequality is shown in figure attached.

5 0

3 years ago

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

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3 years ago

What happens to the value of the expression 3 y over <br> 2 y as y increases

vladimir1956 [14]

It starts becoming bigger :)

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3 years ago

Major arc JL measures 300°. Which describes triangle JLM?

Fed [463]

Answer: equilateral

Step-by-step explanation:

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3 years ago

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