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Anni [7]
3 years ago
11

I really need help .. please help .. the question is in the picture

Mathematics
2 answers:
liberstina [14]3 years ago
6 0

Answer:

C

Step-by-step explanation:

points are on (0,-3) and (6,0)

so

y = 1/2x - 3

tankabanditka [31]3 years ago
5 0

Answer:

C

Step-by-step explanation:

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Tammy predicted that she would sell 300 bottles of water. She actually sold 350 bottles. What is the percent error in the estima
Reil [10]

Answer:

16.67%

Step-by-step explanation:

Step one:

given data

predicted value= 300 bottles

Actual value = 350 bottle

Required:

the % error

Step two:

%error= actual- predicted/predicted *100

%error= 350-300/300 *100

%error= 50/300 *100

%error= 0.166*100

%error= 16.67%

3 0
3 years ago
A slice is made parallel to the base of a square pyramid.
Ivanshal [37]

Answer:

a) Triangle.

b) Triangle.

c) The area is 110 cm.

Step-by-step explanation:

The shape is given because it is a cross section of a Pyramid which we can see shaded in grey.

The area is given by the base times hight devided by 2. So 10cm * 22cm/2= 110cm.

Hope this helped ;)

6 0
3 years ago
Question 24 points)Bill cuts a circle out of a square piece of wood to use for a project.Approximately how much wood will be lef
maks197457 [2]

The square piece of wood measures 32 in per side,

Then the circle that can be cut off it has a radius of 16 in (half of the diameter 32 in)

Therefore, the area of the circle is going to be given by the formula:

\begin{gathered} \text{Area}=\pi R^2 \\ \text{Area}=\pi(16)^2\text{ }\approx803.\, 84in^2 \end{gathered}

And the area of the initial square was:

Area = 32^2 in^2 = 1024 in^2

Therefore the left-overs of the piece of wood would be the difference:

1024 in^2 - 803.84 in^2

3 0
1 year ago
The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC
DIA [1.3K]

Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x²  ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)²  ⇒(2)

Equating (1) and (2)

∴  10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴  10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²

∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

∴ ΔPAQ is a right triangle at A,

PA = 2sqrt5 in  and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

6 0
3 years ago
Ay help its easy pleaseeee.
pickupchik [31]

Answer:

425ft.

Step-by-step explanation:

Just add 395 and 30 together.

Hope this helps. Pls give brainliest!

5 0
3 years ago
Read 2 more answers
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