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3 years ago

Y = x2 + 5x - 14 in factored form need answe asap please

Mathematics

1 answer:

kompoz [17]3 years ago

6 0

Answer:

y = ( x-2)(x+7)

Step-by-step explanation:

y = x^2 + 5x - 14

What 2 numbers multiply to -14 and add to 5

-2 * 7 = -14

-2+7 = 5

y = ( x-2)(x+7)

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Bert is planning to open a savings account that earns 1.6% simple interest yearly. He wants to earn exactly $240 in interest aft

RoseWind [281]

Answer:

D. $5,000

Step-by-step explanation:

The amount of money he should deposit is the principal.

The principal P can be gotten by

P = 100 I /RT

Where

I = interest

R = rate

T = Time

Given

I = $240

R = 1.6%

T = 3 yrs

P = 100 x 240 /1.6 x 3

Multiply through

= 24000/4.8

= $5000

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3 years ago

Rewrite the function by completing the square. h(x)= 4 x^{2} -36 x +81

noname [10]

Answer:

4(x-4.5)^2

Step-by-step explanation:

h(x)=4x^2-36x+81

=4(x^2-9x)+81

=4(x^2-9x+20.25)-(4)(20.25)+81

=4(x-4.5)^2

6 0

3 years ago

alexandera worked 5 hours one day 4.5 hours another and 2.5 another. Find the mean average deviation.

lara31 [8.8K]

Answer:

1.16667

Step-by-step explanation:

3 0

4 years ago

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means

I = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

6 0

3 years ago

Solve x: 6 - 4x = -38 --Please help! This is homework due in for Monday! Xx

igomit [66]

Isolate the x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS (Parenthesis, Exponents (& roots), Multiplication, Division, Addition, Subtraction)

First, subtract 6 from both sides

-4x + 6 (-6) = -38 (-6)

-4x = -38 - 6

-4x = -44

Next, fully isolate the x by dividing -4 from both sides

(-4x)/-4 = (-44)/-4

x = -44/-4

x = 11

11 is your answer for x.

5 0

3 years ago

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