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Harlamova29_29 [7]

2 years ago

12

What is the length of JL?

1 answer:

frozen [14]2 years ago

6 0

As We can see that point J is on (-2,5) and L is on (-2,-2)

So we know that distance cannot be in negative, thus

Distance between J and L =5+2=7 units

JL=7

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Need the answer on this

denpristay [2]

None of the answers are correct because it’s definitely not “Given” because x=20 is like an answer. It’s also not “Distributive Property”, “subtraction Property”, and or “Additional Property” b/c there’s nothing that can be distributed, subtract or add. It’s just an answer.

3 0

2 years ago

Read 2 more answers

Miguel is posting photos on a school website. He has 48 photos, and he is allowed to

weeeeeb [17]

Answer:

12p=48

Step-by-step explanation:

8 0

2 years ago

How I can solve this

gavmur [86]

Answer:

$\boxed{\textrm{B)}\qquad y>4}.$

Step-by-step explanation:

We start by subtracting $1$ from both sides of the second inequality:

$2x > 5 \iff 2x - 1 > 5 - 1 \iff 2x - 1 > 4.$

Therefore, we conclude that:

$y > 2x - 1 > 4 \implies y > 4.$

Therefore, every point that satisfies the system has a $y$ -coordinate greater than $4$ .

7 0

3 years ago

PLEASE HELP ME D: given the parent function of f(x) = x3, what change will occur when the function is changed to f(x − 3)

Alexxandr [17]

Hello,

Your answer would be:

The parent function is : f ( x ) = x³
f ( x + 3 ) = ( x + 3 )³
shift to the left 3 units.

Plz mark me brainliest!

Hope this helps!

4 0

3 years ago

Find the area of the right triangle △DEF with the points D (0, 0), E (1, 1), and F.

Tems11 [23]

Answer:

The area of the triangle is $\sqrt{3}$

Step-by-step explanation:

Given:

Coordinates D (0, 0), E (1, 1)

Angle ∠DEF = 60°

△DEF is a Right triangle

To Find:

The area of the triangle

Solution:

The area of the triangle is = $\frac{1}{2}(base \times height)$

Here the base is Distance between D and E

calculation the distance using the distance formula, we get

DE = $\sqrt{(0-1)^2 + (0-1)^2}$

DE = $\sqrt{(-1) ^2 + (-1)^2$

DE = $\sqrt{1+1}$

DE = $\sqrt{2}$

Base = $\sqrt{2}$

Height is DF

DF = $tan(60^{\circ}) \times DE$

DF = $\sqrt{3} \times DE$

DF = $\sqrt{3} \times\sqrt{2}$

Now, the area of the triangle is

= $\frac{1}{2}({\sqrt{2})(\sqrt{3} \times \sqrt{2})$

= $\frac{1}{2}({\sqrt{2})(\sqrt{3} \sqrt{2})$

= $\frac{1}{2}(2\sqrt{3} )$

= $\sqrt{3}$

6 0

2 years ago