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3 years ago

The difference between seven times a number and 9 is equal to three times the sum of the number and 2. Find the number

Mathematics

1 answer:

murzikaleks [220]3 years ago

8 0

Answer:

Number:3.75

Equation:7 x-9=3(x+2)

Step-by-step explanation:

Let the number be x.

According to the question,

7 x-9=3(x+2)

7 x-9= 3 x+ 6

7 x- 3 x= 9+6

4 x= 15

x=15/4

x=3.75

If you verify the answer you will get,

11.25=11.25

Thank you!

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Verify :(-50) × [37 + (-7) ] = (-50) × 37 + (-50) × (-7) Those who answers quickly will be marked brainliest

kifflom [539]

Answer:

(-50) x 30 = -1850 + 350

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2 years ago

Read 2 more answers

Blake and his brother are participating in their school’s food drive. Blake has collected 6 1/3 pounds of food and his brother h

sergejj [24]

They will need 2 5/12 more pounds to reach their goal.

<h3>Sum and difference of fractions</h3>

Given the following information as shown:

Amount Blake collected = 6 1/3 pounds
Amount his brother collected = 5 3/4 pounds

If their goal is to collect 14 1/2 pounds of food by the end of the month, the remaining food they need to collect will be expressed as:

y =14 1/2 - (6 1/3 + 5 3/4)

y = 14 1/2 - (19/3 + 23/4)

y = 29/2 - (76+69/12)

y = 29/2 - (145/12)

y = 174-145/12

y = 29/12

y = 2 5/12

Hence they will need 2 5/12 more pounds to reach their goal.

Learn more on sum and difference of fractions here: brainly.com/question/24205483

5 0

2 years ago

Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .