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3 years ago

Solve the equation. 2(5c – 7) = 1

Mathematics

2 answers:

Inga [223]3 years ago

5 0

2(5c – 7) = 1

Distribute the 2 into the numbers within the parenthesis

10c-14=1

Add 14 to both sides

10c=15

Divide 10 on both sides

Final Answer: c= 1 1/2 or 1.5

masha68 [24]3 years ago

4 0

<u><em>c=3/2 is the final answer. </em></u>First, you had to divide by two (2) from both sides of equation, and it gave us ⇒ $\frac{2(5c-7)}{2}= \frac{1}{2}$ . And secondly, you had to simplify, and it gave us ⇒ $5c-7=\frac{1}{2}$ . Next, you had to add by seven (7) from both sides of equation, and it gave us ⇒ $5c-7+7=\frac{1}{2}+7$ . Therefore, you can also simplify, and it gave us ⇒ $5c=\frac{15}{2}$ . In fact, you divide by five (5) from both sides of equation, and it gave us ⇒ $\frac{5c}{5}= \frac{12}{2/5}$ . And finally, simplify it's should be the final answer is c=3/2 is the final answer. Hope this helps! And thank you for posting your question at here on brainly, and have a great day. -Charlie

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Marianna [84]

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4 years ago

Let's say we have a triangle with sides called a, b, and c, and that b is the base. Our triangle has a height that we'll call h.

Irina18 [472]

Answer:

16.34 square inches.

Step-by-step explanation:

Given:

Base of triangle, b = 3 inches

Height of triangle, h = 10.89 inches

Question asked:

What is the area of this triangle in square inches?

Round your answer to the nearest hundredths place .

Solution:

As we know,

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As, we have to find the answer to the nearest hundredths place, hence the area of triangle will be 16.34 square inches.

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3 years ago

Write an equation of the line that passes through (2, -3) and is perpendicular to the line y=-2x-3

Natalija [7]

Answer:

$y=\frac{1}{2} (x-2)-3$ or $y=\frac{1}{2} x-4$

Step-by-step explanation:

The perpendicular slope is the opposite-reciprocal of the original slope:

Perpendicular slope of -2 is $\frac{1}{2}$

Write the equation in point-slope form ( $y=a(x-h)+k$ ):

$y=\frac{1}{2} (x-2)-3$

Convert to slope-intercept form if needed:

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3 years ago

Read 2 more answers

Can someone help me

erik [133]

Answer:

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Step-by-step explanation:

To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.

The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...

(2x -12)/2 = arcsin(16/22×sin(24°))

x ≈ 23.206°

Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...

2x -12 = 0

x -6 = 0 . . . . . divide by 2

x = 6 . . . . . . . add 6

Then the range of values of x will be ...

6 < x < 23.206

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* One way to do this is to make use of the law of cosines:

22² = AB² + AC² -2·AB·AC·cos(48°)

16² = AD² + AC² -2·AD·AC·cos(2x-12)

The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...

22² = 2·AB²·(1-cos(48°))

16² = 2·AB²·(1 -cos(2x-12))

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