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Vedmedyk [2.9K]

1 year ago

10

Tim explained a function in words and Paul wrote an equation. Tim: The amount of money in a savings account increases at a rate

of $225 per month. After eight months, the bank account has $4,580 in it. Paul: y−1,400=56(x+26)
Whose function has the smaller y-intercept?
A. Tim’s with a y-intercept of $2,700
B. Paul’s with a y-intercept of $2,856
C. Paul’s with a y-intercept of $2,800
D. Tim’s with a y-intercept of $2,780

1 answer:

masya89 [10]1 year ago

4 0

D .........is the answer

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Lorie is using long division to find the quotient of x^3 + 6x^2 + 5 and x^2 + x -1, as shown at the bottom of this question.

meriva

Answer:

Option (B)

Step-by-step explanation:

x² + x - 1) x³ + 6x² + 0x + 5 ( x + 5

- (<u>x³ + x² - x</u>)

5x² + x + 5

-(<u>5x² + 5x - 5)</u>

-4x + 10

Result : $x+5+\frac{-4x+10}{(x^{2}+x-1)}$

By the comparing the result with the Lorie's solution,

She made a mistake in the last step in which she combined the constant incorrectly.

Therefore, Option (B) will be the answer.

7 0

1 year ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

1 year ago

Osvoldo has a goal of getting at least 30\%30%30, percent of his grams of carbohydrates each day from whole grains. Today, he at

Alchen [17]

Hi there! Osvoldo did not meet his goal.

(In my answering I suppose you mean 30%, 220 grams of carbohydrates and 55 grams of whole grains, if this is incorrect, please let me know)

Today Osvoldo ate 220 grams of carbohydrates.
10 % of 220 is 22 (divide by 10), and therefore 30 % of 220 is 22 * 3 = 66.

If at least 66 grams of Osvoldo's total consumption consisted of whole grains, he would have met his goal. However, he only ate 55 grams of whole grains (which is less than 66), and therefore he did not meet his goal.

8 0

1 year ago

Read 2 more answers

Little Mexican restaurant sells only two kinds of beef burritos: Mucho beef and Mucho Mucho beef. Last week in the restaurant so

Romashka [77]

Step-by-step explanation:

16+22=38

So then youll check out the price thats worth each. 231$ So what is 38 divided by 231$? Thats your answer

7 0

2 years ago

a ball is thrown with a slingshot at a velocity of 110ft/sec at an angle of 20 degrees above the ground from a height of 4.5 ft.

satela [25.4K]

Answer:

$t=2.47\ s$

Step-by-step explanation:

The equation that models the height of the ball in feet as a function of time is

$h(t) = h_0 + s_0t -16t ^ 2$

Where $h_0$ is the initial height, $s_0$ is the initial velocity and t is the time in seconds.

We know that the initial height is:

$h_0 = 4.5\ ft$

The initial speed is:

$s_0 = 110sin(20\°)\\\\s_0 = 37.62\ ft/s$

So the equation is:

$h (t) = 4.5 + 37.62t -16t ^ 2$

The ball hits the ground when when $h(t) = 0$

So

$4.5 + 37.62t -16t ^ 2 = 0$

We use the quadratic formula to solve the equation for t

For a quadratic equation of the form

$at^2 +bt + c$

The quadratic formula is:

$t=\frac{-b\±\sqrt{b^2 -4ac}}{2a}$

In this case

$a= -16\\\\b=37.62\\\\c=4.5$

Therefore

$t=\frac{-37.62\±\sqrt{(37.62)^2 -4(-16)(4.5)}}{2(-16)}$

$t_1=-0.114\ s\\\\t_2=2.47\ s$

We take the positive solution.

Finally the ball takes 2.47 seconds to touch the ground

4 0

2 years ago