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crimeas [40]
3 years ago
11

Can someone help me please is this correct?

Mathematics
1 answer:
STatiana [176]3 years ago
3 0

Answer:

Subtraction.

Step-by-step explanation:

You Subtract 6 on both sides first and then you multiply the denominator which is 11 to the other side of the equal sign.

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If you invest $1600 at 5.25% for 18 months, how much simple interest is earned?
Damm [24]

Interest calculator for a $600 investment. How much will my investment of 600 dollars be worth in the future? Just a small amount saved every day, week, or month can add up to a large amount over time. In this calculator, the interest is compounded annually.

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3 years ago
Someone please help these practice problems are hard
mr Goodwill [35]

Rewrite the equation in slope-intercept form:

9x - 7y = 21

-7y = -9x + 21

y = (9/7)(x) - 3

The y-intercept is where x = 0.

y = (9/7)(0) - 3

y = 0 - 3

y = -3

The x-intercept is where y = 0.

0 = (9/7)(x) - 3

3 = (9/7)(x)

21 = 9x

x = 21/9

x = 7/3

The correct answer would be C. The x-intercept would be at (7/3, 0) and the y-intercept at (0, -3). Hope that helps! :)

7 0
4 years ago
Joyce Mesnic bought an HP laptop computer at Staples for $699. Joyce made a $100 down payment and financed the
s2008m [1.1K]
699-100=$599
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3 0
3 years ago
. Using ratios and the Pythagorean Theorem the approximate width is 23.5 inches and the height is inches.
Sergio039 [100]

Given:

There is a ratio given as 16:9 of width to height and diagonal is 27 iniches

Required:

We need to find the value of height

Explanation:

By ratio

\begin{gathered} \frac{w}{h}=\frac{16}{9} \\  \\ h=\frac{9}{16}w \end{gathered}

where w is width and h is height

by using pythagorean theorem

\begin{gathered} 27^2=h^2+w^2 \\ 729=\frac{81}{256}w^2+w^2 \\  \\ 186624=81w^2+256w^2 \\ w^2=553.78 \\ w=23.5\text{ in} \end{gathered}

to find h

h=\frac{9}{16}*23.5=13.24\text{ in}

Final answer:

height is 13.24 inches

8 0
1 year ago
If A = 1 2 1 1 and B= 0 -1 1 2 then show that (AB)^-1 = B^-1 A^-1<br><br><br> help meeeee plessss ​
Trava [24]

A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}

where det(<em>A</em>) = 1×1 - 2×1 = -1.

B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}

where det(<em>B</em>) = 0×2 - (-1)×1 = 1. Then

B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

On the other side, we have

AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}

and det(<em>AB</em>) = det(<em>A</em>) det(<em>B</em>) = (-1)×1 = -1. So

(AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

and both matrices are clearly the same.

More generally, we have by definition of inverse,

(AB)(AB)^{-1} = I

where I is the identity matrix. Multiply on the left by <em>A </em>⁻¹ to get

A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}

Multiplication of matrices is associative, so we can regroup terms as

(A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}

Now multiply again on the left by <em>B</em> ⁻¹ and do the same thing:

B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}

7 0
3 years ago
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