All categories

2 years ago

If the point (2-3) lies on the graph of the equation 2y=ax-7 then find the value of a

Mathematics

1 answer:

Tom [10]2 years ago

4 0

Answer:

a = 1/2

Step-by-step explanation:

The given equation is :

2y=ax-7 ...(1)

The point (2,-3) lies on the graph of this equation.

We need to find the value of a.

Put x = 2 and y = -3 in equation (1)

2(-3) = a(2) -7

-6 = 2a-7

-6+7 = 2a

2a = 1

a = 1/2

Hence, the value of a is 1/2.

You might be interested in

Suppose the interval [negative 3−3,negative 1−1] is partitioned into nequals=44 subintervals. What is the subinterval length

Kobotan [32]

Answer: Δx = 0.5

Step-by-step explanation:

We have the interval:

[−3, −1]

and we partition it into 4 equal intervals.

first, the range of our interval is equal to the difference between the extremes, this is:

-1 - (-3) = -1 + 3 = 2

Then, if we divide it into 4, we have 4 segments with a range of:

2/4 = 0.5

Then the subinterval delta is 0.5, and the 4 intervals are:

[−3, -2.5], [−2.5, −2], [−2, −1.5], [−1.5, −1]

6 0

2 years ago

Matthew makes 5 apple pies for every 3 blueberry pies last week Matthew made 15 blueberry pies how many apple pies did he make

KIM [24]

Answer:

Step-by-step explanation:

5:3

3*5=15

This means that 5 apple pies times 5.

5*5=25

25:15

3 0

3 years ago

Which one is greater explain why. 8/10 and 0.91

BARSIC [14]

Answer:

0.91 because 8/10 as a decimal is 0.8

Step-by-step explanation:

so 0.91 is bigger than 0.8

7 0

2 years ago

Read 2 more answers

part 2. Find the value of the trig function indicated, use for that Pythagorean theorem to find the third side if you need it.

andreev551 [17]

Answer: $\bold{5)\ \cos \theta=\dfrac{\sqrt{11}}{6}\qquad 6)\ \tan \theta =\dfrac{8}{17}\qquad 7)\ \cos \theta = \dfrac{4}{3}\qquad 8)\ \cos \theta = \dfrac{\sqrt{10}}{10}}$

Step-by-Step Explanation:

Pythagorean Theorem is: a² + b² = c² , where "c" is the hypotenuse

$5)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{3\sqrt{11}}{18}\quad \rightarrow \large\boxed{\dfrac{\sqrt{11}}{6}}$

Note: (15)² + (3√11)² = hypotenuse² → hypotenuse = 18

$6)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{8}{17}\quad =\large\boxed{\dfrac{8}{17}}$

Note: 8² + 15² = hypotenuse² → hypotenuse = 17

$7)\ \tan \theta=\dfrac{\text{side opposite to}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{20}{15}\quad \rightarrow \large\boxed{\dfrac{4}{3}}$

Note: hypotenuse not needed for tan

$8)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{2}{2\sqrt{10}}\quad =\large\boxed{\dfrac{\sqrt{10}}{10}}$