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Elena-2011 [213]

3 years ago

12

HELP PLZ MATH IM FAILING

1 answer:

ehidna [41]3 years ago

6 0

Answer:

Hello! In this picture I marked the new dot for that point and reflected it across the x axis!

The original coordinates are: 5, -3

and the new coordinates are: 5, 3

Hope that helps!

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Use synthetic division to find (x^2+3x–4)÷(x+4)

Jobisdone [24]

Answer:

X^2-x- 8/x+4

Step-by-step explanation:

X+4 becomes a negative 4

Then multiply -4 by the first number which is 1 so -4×1 is -4. Then subtract -4 by the second number 3. This gives you -1 so multiply that by 4 which is -4. Then subtract 4 from the third number -4. So that'll give you -8. Then add an x-value to every number except the very last number as that number will be you remainder.

7 0

3 years ago

Read 2 more answers

Keith ate 2/6 of candy bar. If there were 6 pieces total, what fraction of the candy bar is left?

IgorC [24]

If there were 6 pieces and Keith ate 2 of the pieces, then there would be 4 pieces left.
$\frac{4}{6}$
and when you reduce it it would be
$\frac{2}{3}$

5 0

3 years ago

Read 2 more answers

Mikes plumbing charges 68 for coming to your house and 17 per hour for labor if I paid him 357, how many hours, h, was mike at m

Roman55 [17]

Answer:

b. $68 + 17h = 357$

f. 17 hours

Step-by-step explanation:

Charges for coming to my house = $68 (this will be the constant of the equation)

Amount charged per hour labor = $17

hours = h

Total amount paid to Mike = 357

The equation that represents this situation can be expressed as:

$68 + 17h = 357$

Solve for h

$68 + 17h - 68 = 357 - 68$ (subtraction property of equality)

$17h = 289$

$\frac{17h}{17} = \frac{289}{17}$ (division property of equality)

$h = 17$

Mike was in my house for 17 hours.

5 0

3 years ago

49^x/4-5 = 343^x/3 what does x equal

ratelena [41]

Answer 8 here is why because 34 x8 is 184

3 0

2 years ago

Can someone help me with this problem

Alex_Xolod [135]

Answer:

<h2>A = 20</h2><h2>P = 6√10</h2>

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

A(-3, 0) , B(3, 2)

$AB=\sqrt{(3-(-3))^2+(2-0)^2}=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=\sqrt{4\cdot10}=\sqrt4\cdot\sqrt{10}=2\sqrt{10}$

A(-3, 0), D(-2, -3)

$AD=\sqrt{(-2-(-3))^2+(-3-0)^2}=\sqrt{1^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

AB = CD and AD = BC

The area of a rectangle:

$A=(AB)(AD)$

Substitute:

$A=(2\sqrt{10})(\sqrt{10})=(2)(10)=20$

The perimeter of a rectangle:

$P=2(AB+AD)$

Substitute:

$P=2(2\sqrt{10}+\sqrt{10})=2(3\sqrt{10})=6\sqrt{10}$

4 0

3 years ago