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tatiyna
3 years ago
15

How to graph the equation

Mathematics
1 answer:
navik [9.2K]3 years ago
5 0
First, you need to change it from standard form to slope-intercept form. To do this, you have to add 3x to both sides. Now that y is by itself, divide by 5 on both sides. Your equation should now be y=3x+3. Then, graph the points. Your x-intercept should be at (-5,0) and your y- intercept should be at (0,3).
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Find the intercept(s) of the following equation.<br> y = 2x^2-8
grin007 [14]
X intercepts- (2,0) and (-2,0)
y intercept- (0,-8)
5 0
2 years ago
HELP! WILL GIVE BRAINLIEST! DO NOT TAKE ANSWERS OFF ONLINE, THEY ARE WRONG!!!! The following is an incomplete paragraph proving
ohaa [14]

Answer:

The correct option is B.

Step-by-step explanation:

Given information: AB\parallel DCAB∥DC and BC\parallel ADBC∥AD .

Draw a diagonal AC.

In triangle BCA and DAC,

AC\cong ACAC≅AC (Reflexive Property of Equality)

\angle BAC\cong \angle DCA∠BAC≅∠DCA ( Alternate Interior Angles Theorem)

\angle BCA\cong \angle DAC∠BCA≅∠DAC ( Alternate Interior Angles Theorem)

The ASA (Angle-Side-Angle) postulate states that two triangles are congruent if two corresponding angles and the included side of are congruent.

By ASA postulate,

\triangle BCA\cong \triangle DAC△BCA≅△DAC

Therefore option B is correct

4 0
2 years ago
4c=-40 i don't know this and i need help<br> ill do what i can in return
melamori03 [73]

Answer:

4c=-40

Divide boyh sides by 4

c=-40/4

c=-10

8 0
2 years ago
Read 2 more answers
Which of these triangles are translations of Triangle A?
kondaur [170]

Answer:

A B C D E F are all the same shape

8 0
3 years ago
Read 2 more answers
Lim x--&gt; 0 (e^x(sinx)(tax))/x^2
Tom [10]

Make use of the known limit,

\displaystyle\lim_{x\to0}\frac{\sin x}x=1

We have

\displaystyle\lim_{x\to0}\frac{e^x\sin x\tan x}{x^2}=\left(\lim_{x\to0}\frac{e^x}{\cos x}\right)\left(\lim_{x\to0}\frac{\sin^2x}{x^2}\right)

since \tan x=\dfrac{\sin x}{\cos x}, and the limit of a product is the same as the product of limits.

\dfrac{e^x}{\cos x} is continuous at x=0, and \dfrac{e^0}{\cos 0}=1. The remaining limit is also 1, since

\displaystyle\lim_{x\to0}\frac{\sin^2x}{x^2}=\left(\lim_{x\to0}\frac{\sin x}x\right)^2=1^2=1

so the overall limit is 1.

4 0
3 years ago
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