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dolphi86 [110]
3 years ago
7

Lesson Solve the equation. 13. x = 8 - 12 X=4

Mathematics
1 answer:
ale4655 [162]3 years ago
3 0

Hey there!

Simplify both sides of the equation

8 - 12 = x

Combine your like terms

(8 - 12)

(8 - 12) = - 4

Thus, x = -4

This means your statement/answer is FALSE because x = -4

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

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astra-53 [7]

a=-2, b=3, c=-4, d=5

 abc-d

abc means multiply those 3 together

 so -2 x 3 x -4 = 24

24-5 = 19

answer is 19

4 0
3 years ago
Helppppppppppp please...
Ede4ka [16]
Order from left to right by row
2
3
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3 years ago
One end of a ladder 32 feet long is placed 10 feet from the outer wall of a building that stands on the ground level. How far up
ladessa [460]

Answer:

30 ft

Step-by-step explanation:

This is a classic right triangle problem, where the length of the ladder represents the hypotenuse, where the ladder is lengthwise from the building is the base of the triangle, and what we are looking for is the height of the triangle.  Pythagorean's Theorem will help us find this length.

32^2-10^2=y^2 so

1024 - 100 = y^2 and

y = 30.4 so 30 feet

4 0
3 years ago
In the Midpoint Rule for triple integrals we use a triple Riemann sum to approximate a triple integral over a box B, where f(x,
Tomtit [17]

To approximate the volume with 8 boxes, we have to split up the interval of integration for each variable into 2 subintervals, [0, 1] and [1, 2]. Each box will have midpoint m_{i,j,k} that is one of all the possible 3-tuples with coordinates either 1/2 or 3/2. That is, we're sampling f(x,y,z)=\cos(xyz) at the 8 points,

(1/2, 1/2, 1/2)

(1/2, 1/2, 3/2)

(1/2, 3/2, 1/2)

(3/2, 1/2, 1/2)

(1/2, 3/2, 3/2)

(3/2, 1/2, 3/2)

(3/2, 3/2, 1/2)

(3/2, 3/2, 3/2)

which are captured by the sequence

m_{i,j,k}=\left(\dfrac{2i-1}2,\dfrac{2j-1}2,\dfrac{2k-1}2\right)

with each of i,j,k being either 1 or 2.

Then the integral of f(x,y,z) over B is approximated by the Riemann sum,

\displaystyle\iiint_B\cos(xyz)\,\mathrm dV\approx\sum_{i=1}^2\sum_{j=1}^2\sum_{k=1}^2\cos m_{i,j,k}\left(\frac{2-0}2\right)^2

=\displaystyle\sum_{i=1}^2\sum_{j=1}^2\sum_{k=1}^2\cos\frac{(2i-1)(2j-1)(2k-1)}8

=\cos\dfrac18+3\cos\dfrac38+3\cos\dfrac98+\cos\dfrac{27}8\approx\boxed{4.104}

(compare to the actual value of about 4.159)

4 0
3 years ago
Guys please solve this problem Someone told me the answer is p=11 but didn’t solve the problem. PLEASE
Alexxx [7]

Answer:

So you have to interior angles so let's do the following

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3 0
3 years ago
Read 2 more answers
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