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iragen [17]
2 years ago
14

Complete the table of values for the linear relation y = -5x + 10.

Mathematics
1 answer:
musickatia [10]2 years ago
3 0

the answer will be -5

Step-by-step explanation:

because you have to minus 5 and 10

so the answer will be x =- 5

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PLEASE HELP I WILL GIVE BRAINLEST AND 20 POINTS PLEASEEEE
rewona [7]

Answer:

Question 4: Which equation is parallel to the above equation and passes through the point (35, 30)

y= \frac{5}{7} x + 5  is the correct answer, I found this by inputting the x and y value of the coordinate (35, 30) onto the equation and solving for y-intercept since the slope of all equations is the same (since it's traveling parallel)

30 = \frac{5}{7} (35) + y\\30 = 25 + y\\5 = y

so the equation would be y= \frac{5}{7} x + 5

Question 5: Which equation is perpendicular to the above equation and passes through the point (35, 30)

y=\frac{-7}{5} x+79 is the correct answer, I found this using the same method as before, input coordinate values into the equation and solve for the y-intercept (The only thing changed from the last answer is the opposite reciprocal slope).

30 = \frac{-7}{5}(35) + y \\30 = -49 + y  \\79 = y

so the equation would be y=\frac{-7}{5} x+79

6 0
2 years ago
Find the length of FT¯¯¯¯¯¯¯ A. 77.71 B. 72.47 C. 56.84 D. 49.42
Mkey [24]

Answer:

D, 49.42

Step-by-step explanation:

ΔVFT=180-90-43=47

formula

a/sin A = b/sin B/ = c/sin C

So,

FV/sin90=53/sin47

FV=72.4684

FT=√(72.4684)^2-(53)^2

FT=49.4234

Ans:D

5 0
3 years ago
Read 2 more answers
Find the value of cos 0 , If sec 0 = 3/2 0 < 90
UNO [17]

Answer:

Step-by-step explanation:

cos theta is 2/3 because  cos means reciprocal of sec

7 0
2 years ago
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Please help me, just need the answer and thank you!
Triss [41]
Answer to what? I’m confused lol
6 0
3 years ago
Read 2 more answers
Please please help me
Gekata [30.6K]

Answer:

\large\boxed{\dfrac{\boxed{8}}{81}}

Step-by-step explanation:

\text{If}\ a_1,\ a_2,\ a_3,\ a_4,\ ...,\ a_n\ \text{is the geometric sequence, then}\\\\\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...=\dfrac{a_n}{a_{n-1}}=constant=r-\text{common ration}.\\\\\text{We have}\ \dfrac{1}{2},\ \dfrac{1}{3},\ \dfrac{2}{9},\ \dfrac{4}{27},\ ...\\\\\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{1}{3}\cdot\dfrac{2}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{9}\cdot\dfrac{3}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{4}{27}}{\frac{2}{9}}=\dfrac{4}{27}\cdot\dfrac{9}{2}=\dfrac{2}{3}

\bold{CORRECT}\\\\\dfrac{x}{\frac{4}{27}}=\dfrac{2}{3}\qquad\text{multiply both sides by}\ \dfrac{4}{27}\\\\x=\dfrac{2}{3}\cdot\dfrac{4}{27}\\\\x=\dfrac{8}{81}

8 0
3 years ago
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