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Solnce55 [7]
2 years ago
7

I NEED HELP FAST!!! What is the distance between 14,-15 and 10,-7

Mathematics
1 answer:
kenny6666 [7]2 years ago
4 0

Answer:

8.94 units

Step-by-step explanation:

Distance formula The distance between two points with coordinates

(14,-15) and (10,-7) round to nearest 100th

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What is the difference of the volumes of the two oblique pyramids, both of which have square bases? Round the volumes to the nea
ad-work [718]

Answer:

Step-by-step explanation:

See attachment for the figure

Volume of pyramid can be defined as

V = 1/3 x area of the base x height.

-> Pyramid A:

Volume of Pyramid can be determined by:

V = 1/3 x (2.6cm)² x (2cm) = 4.5067 cm³

Pyramid B:

Volume of Pyramid can be determined by:

V = 1/3 x (2cm)² x (2.5cm) =  3.3333 cm³

Difference b/w two oblique pyramids: 4.5067 cm³ - 3.333 cm³ = 1.17 cm³

By Rounding the volumes to the nearest tenth of a centimeter

1.17cm³ ≈ 1.2cm³

Therefore,  the difference of the volumes of the two oblique pyramids is 1.2cm³

4 0
3 years ago
Rick bought a fan for his living room he was looking at it while he was installing it wondered to himself what the angle between
Nikitich [7]

Answer:

The angle between the two blades is 120 degree.

Step-by-step explanation:

number of blades = 3

The blades are equally spaced.

The total angle around a circle is 360 degree.

So, the angle between the two blades is given by

\theta =\frac{360}{n}\\\theta =\frac{360}{3} = 120^{o}

8 0
3 years ago
Worked 16 3/4 hours in past two days and plan to work 12 3/4 hours in the next two days
Burka [1]

16 3/4 + 12 3/4

16 +12 = 28

3/4 +3/4 = 1 1/2

28 + 1 1/2 = 29 1/2 hours total

8 0
3 years ago
Please help asap!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
vladimir2022 [97]
First one is 42
second is 24
Third is 12
3 0
2 years ago
Read 2 more answers
If someone could help me with number 26...
Volgvan
Given:
\text{sec x} -  \sqrt{\text{2 sec x}-1} =0

Solution:
To solve the equation, it would be best if we remove the root. We remove the root by squaring the equation, but first we need to move the root and the content to the left side.
\text{sec x} -  \sqrt{\text{2 sec x}-1} =0
\text{sec x} =  \sqrt{\text{2 sec x}-1}

Then square both side to remove the root
\text{sec}^2 x= \text{2 sec x}-1

After removing the root, move all terms to the left side
sec² x - 2 sec x + 1 = 0

Do factorization, remember that
a² - 2a + 1 = (a - 1)²
So,
sec² x - 2 sec x + 1 = 0
(sec x - 1)² = 0
sec x - 1 = 0
sec x = 1
\dfrac{1}{\text{cos x}} = 1
cos x = 1
cos x = cos 0°
x = 0°
8 0
3 years ago
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