Ession r' - (3+2) (3 + 2)2 for 2 = 4.

Can someone help me in mathe ( the written blue are the answer choices )

Rudik [331]

4) Here both smaller and larger triangle is "isosceles" in nature.
So, sum would be 180 with two equal sides.
66 + BDE + BED = 180
2 BDE = 180 - 66
BDE = 114 / 2
BDE = 57

In short, Every four angles would be equal to 57

5) A) 7
It is because it must be smaller than the longest side which is equal to 8

6) D) Yes, Because ∠C is congruent to ∠BED

Hope this helps!

6 0

4 years ago

At a carry-out pizza restaurant, an order of 3 slices of pizza, 4 breadsticks, and 2 juice drinks costs $13.35. A second order o

BartSMP [9]

Answer: Pizza = $2.95, Breadstick = $0.50, Juice = $1.25

Step-by-step explanation:

Let P represent the cost of a slice of pizza

and B represent the cost of breadstick

and J represent the cost of a juice drink.

EQ1: 3P + 4B + 2J = 13.35

EQ2: 5P + 2B + 3J = 19.50

EQ3: 4B + J = P + 0.30 --> P - 4B - J = -0.30

Let's eliminate B from EQ1 and EQ2 to form EQ4:

3P + 4B + 2J = 13.35 → 1(3P + 4B + 2J = 13.35) → 3P + 4B + 2J = 13.35

5P + 2B + 3J = 19.50 → -2(5P + 2B + 3J = 19.50) → -10P - 4B - 6J = -39.00

EQ4: -7P -4J = -25.65

And eliminate B from EQ1 and EQ3 to form EQ5:

3P + 4B + 2J = 13.35

P - 4B - J = -0.30

EQ5: 4P + J = 13.05

Now, eliminate J from EQ4 and EQ5 to solve for P:

-7P - 4J = -25.65 → 1(-7P - 4J = -25.65) → -7P - 4J = -25.65

4P + J = 13.05 → 4(4P + J = 13.05) → 16P +4J = 52.20

9P = 26.55

÷9 ÷9

P = 2.95

Plug in P = 2.95 into EQ4 or EQ5 to solve for J:

EQ5: 4P + J = 13.05

4(2.95) + J = 13.05

11.80 + J = 13.05

J = 1.25

Plug in P = 2.95 and J = 1.25 into either EQ1 or EQ2 or EQ3 to solve for B:

EQ3: 4B + J = P + 0.30

4B + 1.25 = 2.95 + 0.30

4B + 1.25 = 3.25

4B = 2.00

B = 0.50

Check (since we used EQ3 to find B, use either EQ1 or EQ2):

EQ2: 5P + 2B + 3J = 19.50

5(2.95) + 2(0.50) + 3(1.25) = 19.50

14.75 + 1.00 + 3.75 = 19.50

19.50 = 19.50 $\checkmark$

6 0

3 years ago

STOP PROCRASTINATING AND DO YOUR WORK

Hitman42 [59]

Answer:

ok jeez you sound like my mom lol

Step-by-step explanation:

4 0

3 years ago

A conical vessels with base radius 5 cm and height 24cm is full of water this water is emptied into a cylindrical vessel of base

Aliun [14]

Answer:

$2\ cm$

Step-by-step explanation:

$We\ are\ given\ that,\\Radius\ of\ the\ conical\ vessel=5\ cm\\Height\ of\ the\ conical\ vessel=24\ cm\\Hence,\\As\ we\ know\ that,\\Volume\ of\ a\ cone=\frac{1}{3}\pi r^2h\\Hence,\\The\ volume\ of\ the\ conical\ vessel=\frac{1}{3}\pi (5)^224= \frac{1}{3}\pi 25*24=200\pi \\Hence,\\The\ volume\ of\ the\ conical\ vessel=The\ amount\ of\ water\ filled\ in\ it\\Hence,\\The\ amount\ of\ water\ filled\ in\ the\ conical\ vessel=200\pi\\Lets\ consider\ the\ cylindrical\ vessel:\\Volume\ of\ a\ cylinder=\pi r^2h$

$As\ we\ are\ not\ said\ that\ the\ cylindrical\ vessel\ was\ empty,\ let\ its\ initial\\ height\ be\ h_1\\Radius\ of\ the\ Cylindrical\ vessel=10\ cm\\Initial\ height\ of\ the\ cylindrical\ vessel=h_1\\Initial\ volume\ of\ the\ cylindrical\ vessel=\pi *10^2*h_1=100h_1\pi \\Initial\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi \\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=Final\ amount\ of\ water\\ in\ the\ cylindrical\ vessel +Amount\ of\ water\ in\ conical\ vessel$

$Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi +200\pi \\Now,\\Let\ the\ final\ height\ be\ h_2\\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=\pi *10^2*h_2=100h_2\pi \\Hence,\\100h_2\pi =100h_1\pi +200\pi \\Hence,\\100h_2=100h_1+200\\100h_2-100h_1=200\\100(h_2-h_1)=200\\h_2-h_1=2\\Hence,\\As\ the\ rise\ in\ level\ is\ the\ difference\ between\ the\ initial\ and\ final\\ heights:\\The\ rise\ in\ level\ of\ water=2\ cm$

8 0

3 years ago

Evaluate 12.5p + 11.7r when p = 6 and r= 7.

Flura [38]

Answer:

12.5*(6) + 11.7*(7)

= 75 + 81.9

= 156.9

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers