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Aleks04 [339]
3 years ago
13

Please help me I need to know the surface area of these 6 shapes 6.

Mathematics
1 answer:
alukav5142 [94]3 years ago
7 0

Answer:

You are correct that the cross section is a rectangle, because a rectangular pyramid has 4 sides connected to the top of the rectangle.

Step-by-step explanation:

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Solve the expression (7th grade math):<br><br> -0.8m+3-12.5m+95.36m
Lana71 [14]

Step-by-step explanation:

<u>Combine Like-Terms:</u>

-0.8m + 3 - 12.5m + 95.36m

-13.3m + 3 + 95.36m

82.06m + 3

82.06m + 3 is your simplified expression.

7 0
2 years ago
What gradient is shown in the graph
Lelu [443]

Answer:

-2/1 or -2

Step-by-step explanation:

in an equation this would be y=-2x-3

5 0
2 years ago
Make t as the subject!! HELP MEE
Umnica [9.8K]

Answer:

t = (p^2/mn) - 1/n

Step-by-step explanation:

Here, we want to make t the subject of the formula

we start by equating both sides so as to remove the root

we have this as;

m(t + n)/t = p^2

m(t + n) = tp^2

mt + mn = tp^2

mn = tp^2 - mt

mn = t(p^2-m)

t = (p^2 - m)/mn

t = p^2/mn - 1/n

6 0
3 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
I WILL AWARD BRAINLIEST TO WHO GETS IT CORRECT A pedestrian is walking at a speed of 3km/h. find the distance the pedestrian wal
ludmilkaskok [199]

Answer:

3km

Step-by-step explanation:

If they are walking 3km for every hour that they walk, then in 1 hour they will walk 3km.

5 0
3 years ago
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