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natka813 [3]
3 years ago
14

How do you solve #5b = - 235#?

Mathematics
1 answer:
NeX [460]3 years ago
4 0
I assume you want to solve this equation for b.  If that is the case, merely divide both sides of this equation by 5:

5b = -35
---     ----
 5        5

Then b = -7 (answer)
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Y = x2 + 4<br> y = x + 6
Nataliya [291]

Answer:

X=2

Y=8

Step-by-step explanation:

Since both equations are equal to y we can equal both equations

2x+4=x+6

x=2

y=2(2)+4

y=8

4 0
3 years ago
Will an exponential function always, never, or sometimes exceed a polynomial function?
patriot [66]
At some point, the exponential function always exceeds a polynomial function.
3 0
3 years ago
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What is the area of this figure?
Elanso [62]
First of all we can draw a parallel line to divide the figure into a triangle and a rectangle as shown in the figure. To find the area of our rectangle, remember that the area of a rectangle is length times width, so A_{r} =lw. Since we know for our figure that the length and width of our rectangle are 13cm and 6cm respectively, lets replace those values in our formula to get its area:
A _{r} =(13cm)(6cm)
A=78cm^{2}
Similarly, the area of a triangle is one half times base times height, so At=( \frac{1}{2})bh. Since we know that our base is 8cm and our height 6cm, lets replace those values in our equation to find the area of our triangle:
A_{t} =( \frac{1}{2})(8cm)(6cm)
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4 0
3 years ago
Helllohello help please if you want -
IRINA_888 [86]

Hey Hey! It's me again!

Answer:

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8 0
3 years ago
Given: ABCD is a ∥-gram, BF ⊥ CD , BE ⊥ AD, Prove: △ABE∼△CBF
drek231 [11]

Answer:

Hence proved △ABE∼△CBF.

Step-by-step explanation:

Given,

ABCD is a parallelogram.

BF ⊥ CD    and

BE ⊥ AD

To Prove : △ABE∼△CBF

We have drawn the diagram for your reference.

Proof:

Since ABCD is a parallelogram,

So according to the property of parallelogram opposite angles are equal in measure.

\therefore m\angle A = m\angle B ⇒1

And given that BF ⊥ CD and BE ⊥ AD.

So we can say that;

m\angle F=m\angle E=90\° ⇒2

Now In △ABE and △CBF

∠A = ∠C   (from 1)

∠E = ∠F    (from 2)

So by A.A. similarity postulate;

△ABE∼△CBF

7 0
3 years ago
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