You can calculate it using the law of cosines: c^2=a^2+b^2-2*a*b*cos(C)

your triangle is
CD=15=a
CE=?=b
DE=CE+3=b+3=c
and C=90°

-> insert those values, with c substituted with b+3 to remove c

c^2=a^2+b^2-2*a*b*cos(C)
(b+3)^2=15^2+b^2-2*15*b*cos(90)
cos(90)=0->
(b+3)^2=15^2+b^2
b^2+2*3*b+3^2=225+b^2
6b+9=225
6b=216
b=36=CE

DE=CE+3=36+3=39

4 0

3 years ago

What is the average rate of change for the function over the interval (-6,-4)

kozerog [31]

Answer:

i believe it is -4 1/3 i may be wrong

Step-by-step explanation:

difficult, let me see....

1st slope:-4

2nd slope:4/6

3rd slope:-3

-2+4/7-3=-5+4/6=-4 2/6=

-4 1/3

5 0

2 years ago

Farmer Jones, and his wife, Dr. Jones, decide to build a fence in their field, to keep the sheep safe. Since Dr. Jones is a math

NISA [10]

Answer: $A=\frac{14}{9}\cdot \sqrt{2}\ unit^2$

Step-by-step explanation:

Given

Two curves are given

$y=10 x^2$

and $y=x^2+2$

the two curves intersect at

$10x^2=x^2+2$

$x=\pm \frac{\sqrt{2}}{3}$

to get the we need to integrate the curves over x axis

$A=\int_{-\frac{\sqrt{2}}{3}}^{\frac{\sqrt{2}}{3}}\left ( x^2+3-10x^2\right )dx$

$A=2\int_{0}^{\frac{\sqrt{2}}{3}}\left ( 3-9x^2\right )dx$

$A=2\left [ 3\left ( \frac{\sqrt{2}}{3}\right )-\frac{9}{3}\left ( \frac{\sqrt{2}}{3}\right )^3\right ]$

$A=2\sqrt{2}\left [ 1-\frac{2}{9}\right ]$

$A=\frac{14}{9}\cdot \sqrt{2}\ unit^2$

3 0

2 years ago

Round 922 to the nearest hundred

artcher [175]

900 because just like rounding tens if you have 14 to 11 itll round to 10 if it is from 15-19 youll round to 20. so since 922 isnt 950-999 youll round it to 900

4 0

2 years ago