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3 years ago

HELP ME ILL GIVE BRAINLIST

Mathematics

1 answer:

Yuki888 [10]3 years ago

4 0

Answer:

y=x and y=2/3x

Step-by-step explanation:

Every point on line a has an equal x and y value so y=x. For line b, the y value of every point is 2/3 of the x value. So at (3,2) 2/3 of 3(the x value) is 2(the y value).

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solong [7]

For these questions it's asking you to substitute the value into the equation.
For example question 5 says,
26= y-9;y=35
basically it's saying if y=35, is the equation correct or not.
In this case: 26= 35 -9 is correct.
So your answer is true.
Hope this helps Xx

7 0

3 years ago

Find the value of x that makes lines u and v parallel.

Temka [501]

You set both expressions equal to each other, then solve for x. Your answer will be x=12

7 0

3 years ago

Read 2 more answers

What is a simplified eCarlotta thinks that 3y + 5y3 is the same as 8y4. Which statement shows that it is NOT the same?Xpression

stira [4]

Questions:

(a) What is a simplified expression for (x + y) – (x – y)?

(b) Carlotta thinks that $3y + 5y^3$ is the same as $8y^4$ . Which statement shows that it is NOT the same?

Answer:

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

$3y + 5y^3 \ne 8y^4$

Step-by-step explanation:

Solving (a):

Given

$(x + y) - (x - y)$

Required

Simplify

$(x + y) - (x - y)$

Open brackets

$(x + y) - (x - y) = x + y - x + y$

Collect Like Terms

$(x + y) - (x - y) = x - x+ y + y$

$(x + y) - (x - y) = y + y$

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

Solving (b):

Given

$3y + 5y^3$ and $8y^4$

Required

Which expression shows they are not the same

The expression that shows this is:

$3y + 5y^3 \ne 8y^4$

Take for instance; y=2

Substitute 2 for y in $3y + 5y^3 \ne 8y^4$

$3*2 + 5*2^3 \ne 8*2^4$

Evaluate all exponents

$9 + 5*8 \ne 8*16$

$9 + 40 \ne 128$

$49 \ne 128$

The above expression supports $3y + 5y^3 \ne 8y^4$

5 0

3 years ago

What best describes the expression y over 4? Some number divided by 4 Some number times 4 4 more than some number 4 less than so

Serhud [2]

y over 4 means the number (y) divided by 4.

5 0

4 years ago

Read 2 more answers

F(x)=(x+1)^(5)-5x-2 Find F'(x) and zeros of F'(x)

KiRa [710]

Assuming you mean
$f(x)=(x+1)^5-5x-2$

first use chain rule on first part
derivitive of (x+1)⁵=5(x+1)⁴ times the derivitive of (x+1) which is 1
the first part is 5(x+1)⁴

derivitive of -5x is -5
derivitive of -2 is 0

we gots
f'(x)=5(x+1)⁴-5
find zeroes
0=5(x+1)⁴-5
5=5(x+1)⁴
1=(x+1)⁴
+/-1=x+1
-1+/-1=x
-2=x or 0=x

f'(x)=5(x+1)⁴-5
zeroes at x=-2 and x=0

3 0

3 years ago