Ask question

Ask question

All categories

Ira Lisetskai [31]

3 years ago

10

Find the value of x for which l || m?

1 answer:

tekilochka [14]3 years ago

8 0

I think the answer for the x will be 37.4 degrees. Hope it help!

You might be interested in

HELP ASAP<br> G = a(4 – d)

NemiM [27]

Answer:

g/-d+4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the volume make it an improper fraction then a mixed number please do a step by step explanation and I will give you brainl

Andrews [41]

Answer:

8463/32=264 15/32

Step-by-step explanation:

Let's make every fraction improper first.

7 3/4=31/4

5 2/8=42/8

6 1/2=13/2

Simplify 42/8

42/8=21/4

21/4

31/4

13/2=26/4

21/14*31/4*26/4=(21*31*26)/64=16926/64=8463/32

3 0

2 years ago

Jimmy is selling bracelets for $1.50. So far she has earned $120. She needs to earn more than $135 in order to meet her goal. Ho

Tema [17]

Selling 90 would reach her goal of $135 so she would need to sell 91 to exceed her goal

135/1.5=90 so 91*1.5=136.5

4 0

3 years ago

Help! please im not sure

loris [4]

The answer is a.
I hope this helped

3 0

3 years ago

Read 2 more answers

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

Read 2 more answers