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2 years ago

Please help me!

Mathematics

1 answer:

rewona [7]2 years ago

6 0

Answer:

12(j+t) + 15 = c

Step-by-step explanation:

So, lets go over what we know.

First off, for simplicity, lets consider Johnathan as J, and Trina as T.

We know that both J and T produce 12 dollars each per hour.

We know that J gets a extra 15 dollars.

We can think of the amount they produce each hour as the time they work in total, times the money they make a hour, which is 12.

We can think of this first part where the total time they work is (J + T)

The 12 dollars they get an hour can be though of as multiplying this sum: 12(J+T)

Finally, we know that J gets an extra 15 dollars.

This is not dependent on time, like the other variables, so we simply add it to the sum of what they make in the end:

12(J+T)+15

This looks like answer C.

Answer:

Hope this helps!

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A rectangle is drawn so the width is 7 inches longer than the height. If the rectangle’s diagonal measurement is 13 inches, don’

cupoosta [38]

Answer:

The height of rectangle is 5 inches

Step-by-step explanation:

The correct question is

A rectangle is drawn so the width is 7 inches longer than the height. If the rectangle’s diagonal measurement is 13 inches, Find the height

Let

x -----> the width of the rectangle in inches

y ----> the height of the rectangle in inches

d ---> diagonal measurement of the rectangle in inches

we know that

Applying the Pythagorean Theorem

$d^2=x^{2}+y^{2}$

we have

$d=13\ in$

$13^2=x^{2}+y^{2}$

$169=x^{2}+y^{2}$ ----> equation A

$x=y+7$ ---> equation B

substitute equation B in equation A

$169=(y+7)^{2}+y^{2}$

solve for y

$169=y^2+14y+49+y^{2}$

$2y^2+14y+49-169=0$

$2y^2+14y-120=0$

solve the quadratic equation by graphing

using a graphing tool

The solution is y=5

see the attached figure

therefore

The height of rectangle is 5 inches

5 0

3 years ago

GET BRAINLIEST FOR THE CORRECT ANSWER AND EXPLANATION!

Tresset [83]

Interpreting the inequality, it is found that the correct option is given by F.

------------------

The first equation is of the line.
The equal sign is present in the inequality, which means that the line is not dashed, which removes option G.

In standard form, the equation of the line is:

$x + 2y = 6$

$2y = 6 - x$

$y = -0.5x + 2$

Thus it is a decreasing line, which removes options J.

We are interested in the region on the plane below the line, that is, less than the line, which removes option H.

------------------

As for the second equation, the normalized equation is:

$3x^2 + 3y^2 = 12$

$3(x^2 + y^2) = 12$

$x^2 + y^2 = 4$

Thus, a circle centered at the origin and with radius 2.
Now, we have to check if the line $x + 2y - 6 = 0$ , with coefficients $a = 1, b = 2, c = -6$ , intersects the circle, of centre $x = 0, y = 0$
First, we find the following distance:

$d = \sqrt{\frac{|ax + by + c|}{a^2 + b^2}}$

Considering the coefficients of the line and the center of the circle.

$d = \sqrt{\frac{|1(0) + 2(0) - 6|}{1^2 + 2^2}} = \sqrt{\frac{6}{5}} = 1.1$

This distance is less than the radius, thus, the line intersects the circle, which removes option K, and states that the correct option is given by F.

A similar problem is given at brainly.com/question/16505684

3 0

3 years ago

Graph the function. H(x) = 6 • (4/3)^x

Aloiza [94]

Answer:

Step-by-step explanation: