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2 years ago

-3x = -12 Solve for x

Mathematics

2 answers:

Trava [24]2 years ago

8 0

Answer:

x = 4

Step-by-step explanation:

Divide both sides of the equation by the same term -3 - -3 and then -12 - -3 then simplify: Cancel terms that are in both the numerator and denominator

and Divide the numbers.

polet [3.4K]2 years ago

3 0

Answer:

Divide each term in

− 3 x = −12 by −3 and simplify. Then ur answer is x =4

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The larger of two numbers is four more than the smaller. if the larger is x what is the smaller?

Lelu [443]

Answer:

x-4

Step-by-step explanation:

Since the larger of two numbers is four more than the smaller one, this means that there is a difference of 4 between the two numbers. In order to find the smaller of both numbers, you can obtain the smaller from the largest by subtracting 4.

4 0

3 years ago

The perimeter of equilateral triangle ABC is 81/3 centimeters, find the length of the radius and apothem.

MAXImum [283]

There is a typo error, the perimeter of equilateral triangle ABC is 81/√3 centimeters.

Answer:

Radius = OB= 27 cm

Apothem = 13.5 cm

A diagram is attached for reference.

Step-by-step explanation:

Given,

The perimeter of equilateral triangle ABC is 81/√3 centimeters.

Substituting this in the formula of perimeter of equilateral triangle = $3\times\ side$

$3\times\ side$ $=[tex]81\sqrt{3}$

$Side = \frac{81\sqrt{3} }{3} =27\sqrt{3} \ cm$

Thus from the diagram , Side $AB=BC=AC= 27\sqrt{3} \ cm$

We know each angle of an equilateral triangle is 60°.

From the diagram, OB is an angle bisector.

Thus $\angle OBC = 30$ °

Apothem is the line segment from the mid point of any side to the center the equilateral triangle.

Therefore considering ΔOBE, and applying tan function.

$tan\theta =\frac{perpendicular}{base} \\tan\theta=\frac{OE}{BE} \\tan\theta=\frac{OE}{\frac{27\sqrt{3}}{2} } \\tan30\times {\frac{27\sqrt{3} }{2} }= OE\\\frac{1}{\sqrt{3} } \times\frac{27\sqrt{3} }{2} =OE\\$

Thus ,apothem OE= 13.5 cm

Now for radius,

We consider ΔOBE

$cos\theta=\frac{base}{hypotenuse} \\cos30= \frac{BE}{OB} \\Cos30 = \frac{\frac{27\sqrt{3} }{2}}{OB} \\OB= \frac{\frac{27\sqrt{3} }{2}}{cos30} \\OB= \frac{\frac{27\sqrt{3} }{2}}{\frac{\sqrt{3} }{2} } \\OB =27 \ cm$

Thus for

Perimeter of equilateral triangle ABC is 81/√3 centimeters,

The radius of equilateral triangle ABC is 27 cm

The apothem of equilateral triangle ABC is 13.5 cm

4 0

3 years ago

Which of these graphs do not represent constant rates of change? Check all that apply

sweet [91]

The first and last graph

4 0

3 years ago

Find the value of x in each case: Given: Iso. ΔABC, HM ∥DG Find: x, m∠CAB, m∠CBA

AleksAgata [21]

1. Start with ΔCIJ.

∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.

2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So

m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.

3. Consider ΔCKL.

∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.

4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So

m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.

5. ΔABC is isosceles, then angles adjacent to the base are congruent:

m∠KBA=m∠JAB → 222°-8x=205°-7x,

7x-8x=205°-222°,

-x=-17°,

x=17°.

Then m∠CAB=m∠CBA=205°-7x=86°.

Answer: 86°.

3 0

3 years ago

Read 2 more answers

Find the difference between 45 2/3 and 27 1/2 . Express your answer in lowest terms

BARSIC [14]

45 2/3 - 27 1/2 = 18.

8 0

3 years ago