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3 years ago

Hurry Quick, help me do this problem. I don't understand how to do this so please help! (Problem in photo)

Mathematics

1 answer:

matrenka [14]3 years ago

7 0

Answer:

1) The system of equations is $5x+4y=75$ and $x+y=18$

2) The first number is $3$ and the second number is $15$

Step-by-step explanation:

1) Let be "x" the first number and "y" the second number.

Remember that:

a- The word "times" indicates multiplication.

b- A sum is the result of an addition.

c- "Is" indicates this sign: $=$

Then, the sum of 5 times "x" and 4 times "y" is 75, can written as:

$5x+4y=75$

And "The sum of the two numbers is 18" can written as:

$x+y=18$

Therefore, the System of equations is:

$\left \{ {{5x+4y=75} \atop {x+y=18}} \right.$

2) You can use the Elimination Method to solve it:

- Multiply the second equation by -5, add the equations and then solve for "y":

$\left \{ {{5x+4y=75} \atop {-5x-5y=-90}} \right.\\......................\\-y=-15\\\\y=15$

- Substitute the value of "y" into any original equation and solve for "x":

$x+15=18\\\\x=18-15\\\\x=3$

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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.

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

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

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

For triangle DEF:

The length of segment DF is to be found,
The length of segment EF is 9,
The sine of angle E is $\sin{64\textdegree}}$ , and
The sine of angle D is $\sin{39\textdegree}$ .

Apply the law of sine:

$\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}$

$\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9$ .

7 0

3 years ago

Triangle abc is shown below

scZoUnD [109]

no it its not there,,......

4 0

3 years ago

Read 2 more answers

Help!<br> C + -328 = 477

krek1111 [17]

The answer for c would be 805

8 0

3 years ago

Read 2 more answers

A woman earns $ 1,350 in interest from two accounts in one year. If she has three times as much invested at 7% as she does at 6%

VashaNatasha [74]

Answer:

The woman invested $15,000 at 7% interest rate and $5,000 at 6% interest rate.

Step-by-step explanation:

We are given the following in the question:

Let x be the interest earned from 7% interest rate and y be the interest earned from 6% interest rate.

The woman invested has three times as much invested at 7% as she does at 6%.

Thus, we can write the equation:

$x = 3y$

The total interest is $1,350.

Thus, we can write the equation:

$1350 = \dfrac{7x}{100} + \dfrac{6y}{100}\\\\7x + 6y = 135000$

Solving the two equations by substitution method:

$7(3y) + 6y = 135000\\27y = 135000\\y = 5000\\x = 3(5000) = 15000$

Thus, she invested $15,000 at 7% interest rate and $5,000 at 6% interest rate.

8 0

3 years ago

What is the factored form of x^2-x-2

tekilochka [14]

For this case we have:

By definition, the factorization is an algebraic expression that is used to find two or more factors, taking into account that the product of these factors must be equal to the given expression.

That is to say:

Given an expression of the form $x ^ 2 + (a + b) x + ab$