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Delvig [45]
3 years ago
13

Find the area for this problem

Mathematics
2 answers:
KIM [24]3 years ago
8 0

Answer:

63 for the whole figure, for the different shaded regions it goes as follows, 17.5 for the white area and 45.5 for the grey area

Step-by-step explanation:

Mamont248 [21]3 years ago
7 0
The answer for the area is 68
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The table shows the low outside temperatures for Monday, Tuesday, and Wednesday.
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-8.8 - (16.5) =
-8.8 + 16.5 =
7.7 <==
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4 years ago
X(2x-5)+x(2x-3)=(4x+6)(x-3)
Helga [31]

Answer:

x=9

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Estimate the distance you can travel in 3 hours 20 minutes if you drive on average 49 miles per hour. Round your answer to the n
AlexFokin [52]

Answer:

163 miles

Step-by-step explanation:

set up a ratio based on 49 miles in 1 hour (60 minutes):

let 'd' = distance in 3hrs, 20 min

3hrs, 20 min = 200 minutes

49/60 = d/200

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7 0
2 years ago
851-473 is the same as ___[dash]-470
Degger [83]
First figure out what 851 minutes 473 equals to, which is 378


So both sides of the equation should equal to 378

Plus 470 and 378 together

Which equals to 848

So the answer is eight hundred and forty eight
8 0
4 years ago
Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and
Lisa [10]

Answer:

Given :

ABC is a right triangle in which ∠ABC = 90°,

Also, Legs AB and CB are extended past point B to points D and E,

Such that,

\angle EAC = \angle ACD = 90^{\circ}

To prove :

EB\times BD=AB\times BC

Proof :

In triangles AEC and EBA,

∠EAC= ∠ABE ( right angles )

∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

\triangle AEC \sim \triangle EBA,

Similarly,

\triangle AEC \sim \triangle ABC

\implies\triangle EBA\sim \triangle ABC-----(1)

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

\triangle ADC \sim \triangle CBD,

Similarly,

\triangle ADC \sim \triangle ABC

\implies \triangle CBD\sim \triangle ABC-----(2)

From equations (1) and (2),

\triangle EBA\sim \triangle CBD

The corresponding sides of similar triangles are in same proportion,

\frac{EB}{BC}=\frac{AB}{BD}

\implies EB\times BD=AB\times BC

Hence, proved....

5 0
3 years ago
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