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3 years ago

HELPPPPP PLEASEEEE!!!!!Find the Area of the Rectangle:

Mathematics

1 answer:

tangare [24]3 years ago

7 0

Answer:

Step-by-step explanation:

IM NOT SURE I THINK BUT hope this helps :)

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PLEASE HELP ASAP

lesantik [10]

This is the way you solve it

6 0

3 years ago

Find the equation of the ellipse with the following properties.

zalisa [80]

Equation of an ellipse:

(x-h)²/a² + (y-k)²/b² = 1
Since it passes through the origin (0,0) , then h = k = 0 hence the equation:
(x-0)²/a² + (y-0)²/b² = 1
x²/a² + y²/b² = 1

2a = major axis = 2.|5| + |-5| = 10. then a = 5 and a² = 25
2b = minor axis = 2.|3| + |-3| = 6. then b = 3 and b² = 9

Then the final equation is:
x²/25 + y²/9 = 1

6 0

4 years ago

Solve please help fast

Natalka [10]

Answer:

5. 40

7. 12

8. 52

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

write an equation of the perpendicular bisector of the line segment whose endpoints are (-1,1) and (7,-5)

gladu [14]

Find midpoint of segment.
(-1,1), (7,-5)
$x= \frac{-1+7}{2}=3,y= \frac{1+(-5)}{2}=-2$

Find slope of line which passes through (-1,1) and (7,-5).
$m= \frac{-5-1}{7-(-1)}= \frac{-6}{8}=- \frac{3}{4}$

The slope of perpendicular line is negative reciprocal value of m.
$m_{\perp}=- \frac{1}{m} =- \frac{1}{- \frac{3}{4} } = \frac{4}{3}$

Write quation of line which slope is $\frac{4}{3}$ and passes throug the point (3,-2).
$m=\frac{4}{3},x_1=3,y_1=-2
\\y-y_1=m(x-x_1)
\\
\\y-(-2)=\frac{4}{3}(x-3)
\\
\\y+2=\frac{4}{3}x-\frac{8}{3}
\\
\\3y+6=4x-8
\\4x-3y-14=0$

3 0

3 years ago

Can you help me with this please

Shalnov [3]

Answer:

see below

Step-by-step explanation:

If (x+5) is a factor of p(x), then the value x=-5 will make that factor be zero. The product p(x) of which it is a part will also be zero, so p(-5) = 0.

Since p(-5) = 0 at x=-5, there will be an x-intercept at x=-5.

Dividing something by a factor of that thing will always give a remainder of zero. So, p(x)/(x+5) has a remainder of 0.

3 0

3 years ago