Which statement is true

Jace is a songwriter who collects royalties in his songs whenever they are played in a commercial or a movie. Jace will earn $50

VashaNatasha [74]

Jace's song were played on 8 commercials and 2 movies

<h3>Further explanation</h3>

Simultaneous Linear Equations could be solved by using several methods such as :

<em>Elimination Method</em>
<em>Substitution Method</em>
<em>Graph Method</em>

If we have two linear equations with 2 variables x and y , then we need to find the value of x and y that satisfying the two equations simultaneously.

Let us tackle the problem!

<em>Number of Commercials = C</em>

<em>Number of Movies = M</em>

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<em>Jaces songs were played on 6 more commercials than movies.</em>

$\boxed{C = 6 + M}$ → <em>Equation 1</em>

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<em>Jace will earn $50 every time one of his songs is played in a commercial and he will earn $140 every time one of his songs is played in a movie.</em>

<em>His total earnings in the royalties from all commercials and movies was $680.</em>

$\boxed{50C + 140M = 680}$ → <em>Equation 2</em>

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<em>Equation 1 ↔ Equation 2 :</em>

$50C + 140M = 680$

$50 ( 6 + M ) + 140M = 680$

$300 + 50M + 140M = 680$

$190M = 680 - 300$

$190M = 380$

$M = 380 \div 190$

$M = 2$

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$C = 6 + M$

$C = 6 + 2$

$C = 8$

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<h2>Conclusion:</h2>

The number of commercials and the number of movies in which Jaces songs where played is respectively 8 and 2

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<h3>Learn more</h3>

Perimeter of Rectangle : brainly.com/question/12826246
Elimination Method : brainly.com/question/11233927
Sum of The Ages : brainly.com/question/11240586

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Simultaneous Linear Equations

Keywords: Simultaneous , Elimination , Substitution , Method , Linear , Equations

5 0

3 years ago

On a morning of a day when the sun will pass directly overhead, the shadow of a 60-ft building on level ground is 25 ft long.

AysviL [449]

Answer:

The shadow is decreasing at the rate of 3.55 inch/min

Step-by-step explanation:

The height of the building = 60ft

The shadow of the building on the level ground is 25ft long

Ѳ is increasing at the rate of 0.24°/min

Using SOHCAHTOA,

Tan Ѳ = opposite/ adjacent

= height of the building / length of the shadow

Tan Ѳ = h/x

X= h/tan Ѳ

Recall that tan Ѳ = sin Ѳ/cos Ѳ

X= h/x (sin Ѳ/cos Ѳ)

Differentiate with respect to t

dx/dt = (-h/sin²Ѳ)dѲ/dt

When x= 25ft

tanѲ = h/x

= 60/25

Ѳ= tan^-1(60/25)

= 67.38°

dѲ/dt= 0.24°/min

Convert the height in ft to inches

1 ft = 12 inches

Therefore, 60ft = 60*12

= 720 inches

Convert degree/min to radian/min

1°= 0.0175radian

Therefore, 0.24° = 0.24 * 0.0175

= 0.0042 radian/min

Recall that

dx/dt = (-h/sin²Ѳ)dѲ/dt

= (-720/sin²(67.38))*0.0042

= (-720/0.8521)*0.0042

-3.55 inch/min

Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min

8 0

4 years ago

Help please!

bagirrra123 [75]

(d): y = mx+n

m = -2/3 ⇒ y = (-2/3)x +n

A(-4, 6) ∈ d ⇒ 6 = (-2/3)·(-4) +n ⇒ 6 = 8/3 +n ⇒

⇒ n = 6 - 8/3 ⇒ n = 10/3

Now, we have:

y = (-2/3)x +10/3

6 0

4 years ago

Evaluate f(t)=-2t2+ 1 for f(-3).

podryga [215]

Answer:

7

Step-by-step explanation:

First, plug-in f(-3) into t in the function:

$f(-3)=-2(-3) +1$