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o-na [289]
3 years ago
12

Brainiliest if u are right 2. Solve the following equation.

Mathematics
2 answers:
Marta_Voda [28]3 years ago
7 0
A is the answer. Look at the attachment for work.

jasenka [17]3 years ago
4 0

Answer:

c

Step-by-step explanation:

7 plus 7 equals 28 and 28 minus 7 equals 21

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7th grade math <br><br><br><br>2 If 40 is decreased by 80%, what is the new amount?​
Afina-wow [57]

Answer:

32

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
3 years ago
Help, with solving rational equations! I am no good at it :( Solve for x: 2 over 5 plus 3 over 5 x equals quantity x plus 5 over
mamaluj [8]
If this is your equation: \frac{2}{5} + \frac{3}{5x} = \frac{x+5}{10}
Solution:
LCD for 5 and 5x is 5x
\frac{x}{x} ( \frac{2}{5} )+ \frac{3}{5x} = \frac{x+5}{10}
\frac{2x}{5x} + \frac{3}{5x} = \frac{x+5}{10}
\frac{2x+3}{5x} = \frac{x+5}{10}
10(2x+3)=5x(x+5) ←cross product
20x + 30 = 5x² + 25x  ←simplify with distributive property
0 = 5x² + 5x - 30 ←use inverse operations to collect all terms on one side
0 = x² + x - 6  ←if possible divide by numerical GCF (This case 5)
0 = (x + 3)(x - 2) ←Factor
x = -3 or x = 2
Please check by substitution in original equation... Both work
4 0
3 years ago
Factorize 2x^3+7x^2-3x-18
Triss [41]
Factor <span><span><span><span>2<span>x3</span></span>+<span>7<span>x2</span></span></span>−<span>3x</span></span>−18</span><span><span><span><span>2<span>x3</span></span>+<span>7<span>x2</span></span></span>−<span>3x</span></span>−18</span><span>=<span><span><span>(<span>x+2</span>)</span><span>(<span>x+3</span>)</span></span><span>(<span><span>2x</span>−3</span>)</span></span></span>Answer:<span><span><span>(<span>x+2</span>)</span><span>(<span>x+3</span>)</span></span><span>(<span><span>2x</span>−3</span><span>)</span></span></span>
5 0
4 years ago
Can someone help me ASAP!!! Please!!!!
g100num [7]

Answer:

step 1. Angle bisector theorem.

step 2. 14/6 = 21/ (3x - 12)

step 3. 14(3x - 12) =21(6)

step 4. 6x - 24 = 18

step 5. 6x = 42

step 6. x = 7.

7 0
3 years ago
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