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bija089 [108]
3 years ago
13

How would I graph a line that contains the point (6,5) and has a slope of - 2/3 ?​

Mathematics
2 answers:
ddd [48]3 years ago
6 0

Answer:

y=-2/3x+9

Step-by-step explanation:

Use point-slope formula:

y-y1 = m(x-x1)

y-5=-2/3(x-6)

y-5=-2/3x+4

y=-2/3x+9

You would then proceed to graphing this equation, as it is in slope intercept form now.

HTH :)

Semenov [28]3 years ago
6 0
Y=-2/3x+5
Answer is
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5/7x =5 <br> Pls pls pls someone help me on this I tried so many times and I can’t figure this out.
Talja [164]

Step-by-step explanation:

\frac{5}{7} x = 5 \\ x =  \frac{5 \times 7}{5}  \\ x = 7

value of x is 7

hope this answer helps you dear....take care!

8 0
2 years ago
Read 2 more answers
Find the lcm of 2.5, 1.0 and 70​
Ivahew [28]

Answer:

Step-by-step explanation:

2.5 = 5 * .5

1 = 1

70 = 2 * 5 *  7

LCM = 2 * 5 * 7

If you include the 1/2, you will reduce the LCM to 35, but 70 will be left out of the LCM.

4 0
2 years ago
How do you find the hypotenuse of a right triangle when you have 2 angles (1 being the 90 degrees since it is a right triangle)
stiks02 [169]
Find the length of the hypotenuse by using either "sine" or "cosine". 

Sine= Length of opposite side/ Hypotenuse

Cosine= Length of adjacent side/ Hypotenuse

In order to use one of these, it depends on which angle they give you other than 90 degrees. 

Hope this helps!
8 0
3 years ago
Whoever answers first can get brainliest!
Andrew [12]

Answer:

d

Step-by-step explanation:

8 0
3 years ago
Find the quotient of these Complex Numbers. <br><br><br> (5-i) / (3+2i)
JulsSmile [24]

Let the given complex number

z = x + ix = \dfrac{5-i}{3+2i}

We have to find the standard form of complex number.

Solution:

∴ x + iy = \dfrac{5-i}{3+2i}

Rationalising numerator part of complex number, we get

x + iy = \dfrac{5-i}{3+2i}\times \dfrac{3-2i}{3-2i}

⇒ x + iy = \dfrac{(5-i)(3-2i)}{3^2-(2i)^2}

Using the algebraic identity:

(a + b)(a - b) = a^{2} - b^{2}

⇒ x + iy = \dfrac{15-10i-3i+2i^2}{9-4i^2}

⇒ x + iy = \dfrac{15-13i+2(-1)}{9-4(-1)} [ ∵ i^{2} =-1]

⇒ x + iy = \dfrac{15-2-13i}{9+4}

⇒ x + iy = \dfrac{13-13i}{13}

⇒ x + iy = \dfrac{13(1-i)}{13}

⇒ x + iy = 1 - i

Thus, the given complex number in standard form as "1 - i".

5 0
3 years ago
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