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3 years ago

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has vertices J(3, –1), K(4, –4), and L(1, –3). Determine if is an equilateral, isosceles, or scalene triangle. A. equilateral B.

isosceles C. scalene D. none of these

2 answers:

Juli2301 [7.4K]3 years ago

7 0

Answer c) scalene :-) ....

Crazy boy [7]3 years ago

3 0

Answer:

B. Isosceles

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

Obtain the general solution to the equation. (x^2+10) + xy = 4x=0 The general solution is y(x) = ignoring lost solutions, if any

alukav5142 [94]

Answer:

$y(x)=4+\frac{C}{\sqrt{x^2+10}}$

Step-by-step explanation:

We are given that a differential equation

$(x^2+10)y'+xy-4x=0$

We have to find the general solution of given differential equation

$y'+\frac{x}{x^2+10}y-\frac{4x}{x^2+10}=0$

$y'+\frac{x}{x^2+10}y=4\frac{x}{x^2+10}$

Compare with

$y'+P(x) y=Q(x)$

We get

$P(x)=\frac{x}{x^2+10}$

$Q(x)=\frac{4x}{x^2+10}$

I.F= $e^{\int\frac{x}{x^2+10} dx}=e^{\frac{1}{2}ln(x^2+10)}$

$e^{ln\sqrt(x^2+10)}=\sqrt{x^2+10}$

$y\cdot \sqrt{x^2+10}=\int \frac{4x}{x^2+10}\times \sqrt{x^2+10} dx+C$

$y\cdot \sqrt{x^2+10}=\int \frac{4x}{\sqrt{x^2+10}}+C$

$y\cdot \sqrt{x^2+10}=4\sqrt{x^2+10}+C$

$y(x)=4+\frac{C}{\sqrt{x^2+10}}$

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3 years ago

Please help! Write the intervals of the values shown on the graph below

Veseljchak [2.6K]

The answer is
-1 ≤ x < 3
We just learned this is in my algebra 2 AP class

5 0

3 years ago

Help please tell me what the answer is ASAP

Amiraneli [1.4K]

Answer:

The approximated length of EF is 2.2 units ⇒ A

Step-by-step explanation:

<em>The tangent ratio in the right triangle is the ratio between the opposite side to the adjacent side of one of the acute angle in the triangle</em>

In the given figure

∵ The triangle DFE has a right angle F

∵ The opposite side to angle D is EF

∵ The adjacent side to angle D is DF

→ By using the tangent ratio above

∴ tan(∠D) = $\frac{EF}{DF}$

∵ DF = 6 units

∵ m∠D = 20°

→ Substitute then in the ratio above

∴ tan(20°) = $\frac{EF}{6}$

→ Multiply both sides by 6

∴ 6 tan(20°) = EF

∴ 2.183821406 = EF

→ Approximate it to the nearest tenth

∴ 2.2 = EF

∴ The approximated length of EF is 2.2 units

7 0

3 years ago