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Stella [2.4K]
3 years ago
10

What is b-2 = - 12 Ill give brainlist!

Mathematics
2 answers:
Ede4ka [16]3 years ago
5 0

Are you asking for what is B?

Cause B= -14

-14-2 = -12

I forgot when you subtract a negative from a negative number it turns into additionnnn T-T

mash [69]3 years ago
4 0

Answer:

-14

Step-by-step explanation:

-14 - 2 = - 12

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yes was walking along the road at 5 mph and was 17 miles away from his destination. no was walking at 7 mph but was 22 miles awa
yarga [219]
17 miles / 5 mph = 3.4 hours

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No would arrive before yes
7 0
3 years ago
¿Cuántas esferas hay en la figura 20?
scoundrel [369]

Answer:

418

Step-by-step explanation:

fig 2 = fig 1 + 5

fig 3 = fig 2 + 7 or fig 1 +5+7   16

1  2  3   4  5   6    7  8   9  10   11   12   13  14   15   16   17  18  19  20

4+5+7+9+11+13+15+17+18+19+21+23+25+27+29+31+33+35+37+39

8 0
3 years ago
The equation of a line is given below . 6x+2y =-18 find the x intercept and the y intercept
vova2212 [387]

Answer: x intercept is -3

y intercept is -9

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6 0
2 years ago
The polynomial x 3 + 5x 2 - ­57x -­189 expresses the volume, in cubic inches, of a shipping box, and the width is (x+3) in. If t
Ipatiy [6.2K]
V=x^3+5x^2-57x-189
Width: W=(x+3) in = 15 in →x+3=15
Solving for x:
x+3-3=15-3→x=12

With x=12 the Volume would be:
V=(12)^3+5(12)^2-57(12)-189
V=1,728+5(144)-684-189
V=1,728+720-684-189
V=1,575

V=W*D*H
Depth: D
Height: H
with H>D

V=1,575; W=15
Replacing in the equation above:
1,575=15*D*H
Dividing both sides by 15
1,575/15=(12*D*H)/15
105=D*H
3*5*7=D*H
D<H
If D=5→H=3*7→H=21
If D=7→H=3*5→H=15

Answer: Option <span>C. height: 21 in. depth: 5 in.
</span>
Please, see the attached file for another form to solve the problem

4 0
3 years ago
Read 2 more answers
Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x2 from (−1, 2) to (1, 2) (a) find a function f
Korvikt [17]
If there is some scalar function f(x,y) such that

\nabla f(x,y)=\mathbf f(x,y)=x^2\,\mathbf i+y^2\,\mathbf j

then we want to find f such that

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\dfrac{\partial f}{\partial y}=y^2=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\dfrac{y^3}3+C
\implies f(x,y)=\dfrac{x^3}3+\dfrac{y^3}3+C

So the vector field \mathbf f(x,y) is conservative, which means the fundamental theorem applies; the line integral of \mathbf f along any path \mathcal C parameterized by some vector-valued function \mathbf r(t) over a\le t\le b is given by

\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=a}^{t=b}\mathbf f(\mathbf r(t))\cdot\dfrac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt=f(\mathbf r(b))-f(\mathbf r(a))

In this case,

\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(1,2)-f(-1,2)=\dfrac23
5 0
3 years ago
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