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nexus9112 [7]
3 years ago
9

Line A passes through the points (-1, 4) and (4, 9). Line B passes through the points (4, -4) and (-3, 24). Find the point where

line A intersects line B
Mathematics
2 answers:
exis [7]3 years ago
7 0
A calculator can help
iVinArrow [24]3 years ago
6 0

Answer: y= 4x-2

Step-by-step explanation: y=mx+b

(-6)= (4) (-1) +b

-6=-4+b

-2=b

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Karin wants to use the distributive property to mentally find the value of 19⋅42+19⋅58. Which expression can she use?
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3 0
3 years ago
Read 2 more answers
a regular rectangular pyramid has a base and lateral faces that are congruent equilateral triangles. it has a lateral surface ar
Margarita [4]
A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)

Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say x.

The lateral faces are equilateral triangles of side length x.

The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18  cm^2.

now we need to find x. Consider the picture attached, showing one lateral face of the pyramid.

by the Pythagorean theorem: 

h= \sqrt{ x^{2} - (x/2)^{2}}= \sqrt{ x^{2}- x^{2}/4}= \sqrt{3x^2/4}= \frac{ \sqrt{3} }{2}x

thus, 

Area_{triangle}= \frac{1}{2}\cdot base \cdot height\\\\18= \frac{1}{2}\cdot x \cdot \frac{ \sqrt{3} }{2}x\\\\ \frac{18 \cdot 4}{ \sqrt{3}}=x^2

thus:

x^2 =\frac{18 \cdot 4}{ \sqrt{3}}= \frac{18 \cdot 4 \cdot\  \sqrt{3} }{3}=24 \sqrt{3}       (cm^2)

but x^{2} is exactly the base area, since the base is a square of sidelength = x cm.


So, the total surface area = base area + lateral area =  24 \sqrt{3}+72   cm^2


Answer: 24 \sqrt{3}+72   cm^2

4 0
3 years ago
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