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vampirchik [111]
2 years ago
8

Line c has an equation of y=-3/4x-3. Line d is parallel to line c and passes through the points (-2, -2). What is the equation o

f line d in slope intercept form?
Mathematics
1 answer:
blagie [28]2 years ago
5 0

Answer:

y = \frac{3}{4} x - \frac{7}{2}

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - \frac{3}{4} x - 3 ← is in slope- intercept form

with slope m = - \frac{3}{4}

Parallel lines have equal slopes , then

y = - \frac{3}{4} x + c ← is the partial equation of line d

to find c substitute (- 2, - 2 ) into the partial equation

- 2 = \frac{3}{2} + c ⇒ c = - 2 - \frac{3}{2} = - \frac{7}{2}

y = - \frac{3}{4} x - \frac{7}{2} ← equation of line d

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Share 270g in the ratio 5 : 4
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Answer:

150:120

Step-by-step explanation:

U add the 2 ratios 5+4=9

270/9=30

30*5=150

30*4=120

8 0
3 years ago
A nursery has 10 tree seedlings to give out at 2 workshops. It wants to give out a minimum of 2 seedlings at each workshop. How
11111nata11111 [884]

We are given total 10 tree seedlings.

Number of workshops = 2.

Here order doesn't matters.

<em>When order doesn't matters we apply combination.</em>

We know formula of combination:

C(n,r) = \frac{n!}{r!(n-r)!}

For the given situation, we have

n=10 and r=2.

Plugging values in formula, we get

\frac{10!}{2!(10-8)!}  = \frac{10\times 9\times8!}{2!8!}

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<h3>Therefore, 45 different ways can the nursery give out seedlings.</h3>
5 0
3 years ago
What is the minimum value of C = 7x + 8y, given the constraints: 2x + y ≥ 8, x + y ≥ 6, x ≥ 0, y ≥ 0. A. 32 B. 42 C. 46 D. 64
marshall27 [118]

Answer: The minimum value of C is 46.

Step-by-step explanation:

Since, Here, We have to find out Min C = 7x+8y

Given the constraints are 2x+y\geq 8 -------(1)

x+y \geq 6   ------------- (2)

x \geq 0, y \geq 0  -------- (3)

Since, For equation 1) x-intercept, (4, 0) and y-intercept (0,8)

And, 2\times 0+0\geq 8⇒0\geq 8 ( false)

Therefore the area of line 1) does not contain the origin.

For equation 2) x-intercept, (6, 0) and y-intercept (0,6)

And, 0+0\geq 6⇒0\geq 6 ( false)

Therefore the area of line 2) does not contain the origin.

Thus after plotting the constraints 1) 2) and 3) we get Open Shaded feasible region AEB ( Shown in below graph)

At A≡(0,8) , C= 64

At E≡(2,4),  C= 46

At B≡(6,0),  C= 42

Thus at B, C is minimum, And its minimum value = 42


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