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2 years ago

What is 4y - 9x + 3y - 2x when simplified?

Mathematics

1 answer:

pychu [463]2 years ago

3 0

Answer:

A. -11x + 7y

Step-by-step explanation:

4y - 9x + 3y - 2x

4y + 3y = 7y

-9x - 2x = -11x

-11x + 7y

I hope this helps!

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Ricky went to buy stationary to New Paper Mart. He found missing data in the bill as shown below. Find the values of A, B, C and

Bond [772]

Answer:

The value of A = 36

The value of B = 42

The value of C = 30

The value of D = 12

Step-by-step explanation:

A= $6 x 6= 36

B= 336 ÷ 8= 42

C= $15 x 2= 30

D= 24 ÷ $2= 12

Find complete question in the attachment

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3 years ago

(2^8 ⋅ 3^−5 ⋅ 6^0)^−2 ⋅ 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 ⋅ 2^28 (5 points) Write y

mihalych1998 [28]

Start by looking at simplifying the smaller problems

anything to the power zero is 1 $6^0 = 1$

next apply the power of a power rule... multiply the powers

$(2^8\times 3^{-5} \times 1 )^{-2} = 2^{8\times -2}\times 3^{-5 \times -2}$

which becomes

$2^{-16}\times 3^{10}$

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3 years ago

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kogti [31]

It would be HL theorem because the triangles are at 90 deg with a right angle

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3 years ago

Can someone help me with this please?

belka [17]

Blue line y= 2x
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3 years ago

What are the first 10 digits after the decimal point (technically the hexadecimal point...) when the fraction frac17 is written

Nataliya [291]

We happen to have

$\dfrac17 = \dfrac18 + \dfrac1{8^2} + \dfrac1{8^3} + \cdots$

which is to say, the base-8 representation of 1/7 is

$\dfrac17 \equiv 0.111\ldots_8$

This follows from the well-known result on geometric series,

$\displaystyle \sum_{n=1}^\infty ar^{n-1} = \frac a{1-r}$

if $|r|$ . With $a=1$ and $r=\frac18$ , we have

$\displaystyle \sum_{n=1}^\infty \frac1{8^{n-1}} = 1 + \frac18 + \frac1{8^2} + \frac1{8^3} + \cdots \\\\ \implies \frac1{1-\frac18} = 1 + \frac18 + \frac1{8^2} + \frac1{8^3} + \cdots \\\\ \implies \frac87 = 1 + \frac18 + \frac1{8^2} + \frac1{8^3} + \cdots \\\\ \implies \frac17 = \frac18 + \frac1{8^2} + \frac1{8^3} + \cdots$

Uniformly multiplying each term on the right by an appropriate power of 2, we have

$\dfrac17 = \dfrac2{16} + \dfrac{2^2}{16^2} + \dfrac{2^3}{16^3} + \dfrac{2^4}{16^4} + \dfrac{2^5}{16^5} + \dfrac{2^6}{16^6} + \cdots$

Now observe that for $n\ge4$ , each numerator on the right side side will contain a factor of 16 that can be eliminated.

$\dfrac{2^n}{16^n} = \dfrac{2^4\times2^{n-4}}{16^n} = \dfrac{2^{n-4}}{16^{n-1}}$

That is,

$\dfrac{2^4}{16^4} = \dfrac1{16^3}$

$\dfrac{2^5}{16^5} = \dfrac2{16^4}$

$\dfrac{2^6}{16^6} = \dfrac4{16^5}$

etc. so that

$\dfrac17 = \dfrac2{16} + \dfrac4{16^2} + \dfrac9{16^3} + \dfrac2{16^4} + \dfrac4{16^5} + \dfrac9{16^6} + \cdots$

and thus the base-16 representation of 1/7 is

$\dfrac17 \equiv 0.249249249\ldots_{16}$

and the first 10 digits after the (hexa)decimal point are {2, 4, 9, 2, 4, 9, 2, 4, 9, 2}.

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2 years ago