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jasenka [17]
2 years ago
7

Please help me solve this problem

Mathematics
1 answer:
algol [13]2 years ago
7 0
1.afgx2
2.bfgx2

I hope it helps Im really not that smart

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2(3w-5y) expanded <br>please ​
Snezhnost [94]
<h2><u>DISTRIBUTIVE PROPERTY</u></h2>

The distributive property indicates that multiplying a number by the sum of two addends is equal to multiplying each addend individually by the number and then adding the products together.

In general terms:

<h3>m(n + p) = mn + mp</h3>

‎      ‏‏‎

<u>In this exercise</u>:

Let's apply the distributive property:

\mathsf{2(3w - 5y)}

= \mathsf{2(3w) - 2(5y)}

= \boxed{\mathsf{6w - 10y}}

‎      ‏‏‎

<h3><u>Answer</u>: 6w - 10y</h3>

‎      ‏‏‎

7 0
3 years ago
What is the value of (j+3k)(5)?<br><br> j(x) = (x + 4)^2; k(x) = 8 - x<br><br> (j + 3k)(5) =
jeyben [28]

Step-by-step explanation:

J=(x+4)²

K=(8-x)

X=5

=J+3k

=(x+4)(x+4)+3(8-x)

=(x²+8x+16+24-3x)

=25+25+40

=90

Alternatively:

J=(5+4)²

J=81

K=8-5

K=3

J+3k

81+3(3)

81+9

90

3 0
3 years ago
Check all solutions to the equation. If there are no solutions, check "None."x2=0A. 1B. -1C.0D. 10E. 4E None
Juli2301 [7.4K]

The given equation is,

\begin{gathered} x^2=0 \\ x=0 \end{gathered}

Thus, option (C) is the correct solution.

8 0
1 year ago
Which number equals (3)^-3? A) -27 B) -9 C) -1/27 D) 1/27
Alecsey [184]

Answer:

the answer is D 1/27

Step-by-step explanation:

7 0
3 years ago
Find the volume of the solid under the plane 5x + 9y − z = 0 and above the region bounded by y = x and y = x4.
svp [43]
<span>For the plane, we have z = 5x + 9y

For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.

Since x > x^4 for y in [0, 1],

The volume of the solid equals

\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx  \\  \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx  \\  \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\  \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0

=\frac{19}{6} - \frac{5}{6} - \frac{1}{2} =\bold{ \frac{11}{6} \ cubic \ units}</span>
8 0
3 years ago
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