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2 years ago

Subtract. (2p^2+q)-(p^2+q) Is the answer P^2 P^2+2q P^2-2q 3p^2

Mathematics

1 answer:

Fofino [41]2 years ago

4 0

Answer:

-p5q2 - p3q5 + 2p3 + 2pq2 + 2q3

____________________________

p3q2

Step-by-step explanation:

Simplify ——

Equation at the end of step 1 :

(p+q) 2

(((2•—————)-(q3))+——)-p2

(p3) q2

Step 2 :

p + q

Simplify —————

p3 (p + q) 2

(((2 • ———————) - q3) + ——) - p2

p3 q2

Step 3 : 2 • (p + q) 2

((——————————— - q3) + ——) - p2

p3 q2 q3 q3 • p3

q3 = —— = ———————

1 p3

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Pls can someone help me and have it right

dangina [55]

Answer:

Problem 1:

a. x=2

b. x=3

c. x=1

Problem 2:

A multiplication equation to hold the table true:

$18\times3=54$

A division equation to hold the table true:

$\frac{54}{3} =18$

Step-by-step explanation:

Given in problem 1:

(a). The equation is $\frac{x}{6} > 1$

It holds true for all values of $x> 6$ .

Let us say $x=12$ ,

$\frac{x}{6}=\frac{12}{6} =2$ which is greater than 1.

(b). The equation is $\frac{x}{6} < 1$

It holds true for all values of $x< 6$ .

Let us say $x=3$

$\frac{x}{6} =\frac{3}{6} =\frac{1}{2}$ which is less than 1.

(c). The equation is $\frac{x}{6} = 1$

It holds true for only $x= 6$ .

Let us say $x=6$ ,

$\frac{x}{6} =\frac{6}{6}=1$ which is equal to 1.

Problem 2:

A multiplication equation to hold the table true:

$18\times3=54$

A division equation to hold the table true:

$\frac{54}{3} =18$

Therefore these are the values which hold true to the equation in problem 1 and 2.

8 0

2 years ago

Solve the quadratic equation y^2-2y-35=0

soldier1979 [14.2K]

Hello,

To solve we need to factor y^2-2y-35:

(y-7)(y+5)=0

Now solve for :

(When will (y-7)(y+5)equal zero?)

When y-7=0 or y+5=0

Solve for the 2 equations above and;

y= 7,-5

Hope this helps, have a great day!

6 0

2 years ago

How do you solve this -6 ÷[-3/4]=?

sergij07 [2.7K]

The rule to use: when dividing fractions, invert and multiply.

-6 ÷ -(3/4) equals -6 * -(4/3) which equals 24 / 3 which equals

7 0

2 years ago

Read 2 more answers

ERRE

n200080 [17]

Answer:

<em>The total percentage discount was 19%</em>

Step-by-step explanation:

<u>Percentages</u>

Suppose the original price of the shirt was x.

A discount of 10% reduces the price by:

10*x/100 = 0.1x

Thus, the discounted price is:

x - 0.1x = 0.9x

The second discount reduces the price by:

10*0.9x/100 = 0.09x

The new discounted price is:

0.9x - 0.09x = 0.81x

With respect to the original price, the last discounted price has been discounted by:

x - 0.81x = 0.19x

This means the total percentage discount was 0.19*100 = 19%.

The total percentage discount was 19%

5 0

2 years ago

Which statement about the figure MUST be true? Select all that apply.

olga55 [171]

<h2>○=> <u>Correct options</u> :</h2><h2>□ $\color{plum}\tt\bold{(C) \: SR = 27}$ </h2><h2>□ $\color{plum}\tt\bold{(D) \: QR = 54 }$ </h2><h3><u>Steps to derive the correct options</u> :</h3>

Since two sides and one included angle is equal in △PQS and △PRS, we can conclude that they are congruent under the SAS congruence criterion.

Which means :

▪︎Angle S = Angle S

▪︎PS = PS

▪︎QS = RS

Given :

Measure of segment QS = 6n+3

Measure of segment RS = 4n+11

Thus :

$= \tt6n + 3 = 4n + 11$

$= \tt6n + 3 - 4n = 11$

$=\tt 2n + 3 = 11$

$= \tt2n = 11 - 3$

$= \tt2n = 8$

$=\tt n = \frac{8}{2}$

$\hookrightarrow\color{plum}\tt n = 4$

Thus, the value of n = 4

Measure of segment QS :

$=\tt 6n + 3$

$= \tt6 \times 4 + 3$

$= \tt24 + 3$

$\color{plum}\tt \: \bold{QS = 6n + 3 = \tt\bold{27}}$

Thus, measure of QS = 27

Measure of RS :

$=\tt 4n + 11$

$=\tt 4 × 4 + 11$

$=\tt 16+11$

$\color{plum}\tt \: \bold{RS = 4n + 11 = \tt\bold{27}}$

Measure of QR :

$=\tt 27+27$

$\color{plum}\tt \: \bold{QR = 27+27 \tt\bold{=54}}$

Thus :

▪︎QS = 27

▪︎RS = 27

▪︎QR = 54

Therefore, the correct options are :

▪︎(C) SR = 27

▪︎(D) QR = 54

7 0

2 years ago