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3 years ago

ANSWER THIS AND GET BRAINLIEST AND 30 POINTS

Mathematics

2 answers:

lbvjy [14]3 years ago

7 0

Answer:

After t weeks, the balance in the expense account and fundraising account is

f(t)= 400 -40t +150+32t

f(t)= 550 -8t

After 12 weeks,

f(12) = 550- 8(12) = 454

ElenaW [278]3 years ago

6 0

Answer:

Step-by-step explanation:

After t weeks, the balance in the expense account and fundraising account is

f(t)= 400 -40t +150+32t

f(t)= 550 -8t

After 12 weeks,

f(12) = 550- 8(12) = 454

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The prices of two ratios are in the ratio x:y

Archy [21]

Answer:

4:1

Step-by-step explanation:

x+20 : y+20 = 5 : 2

(x+20)/(y+20) = 5/2

2(x+20) = 5(y+20)

2x+40 = 5y+100

2x = 5y+60 (1)

x-5 : y-5 = 5:1

(x-5)/(y-5) = 5/1

x-5 = 5(y-5)

x-5 = 5y-25

x = 5y-20 (2)

Solve (1) & (2) simultaneously,

2(5y-20) = 5y+60

10y-40 = 5y+60

5y = 100

y = 20

x = 5y-20

x = 5(20)-20 = 100-20

x = 80

x:y

80:20

4:1

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3 years ago

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Forty six times y is no more than 276

Step2247 [10]

Answer:

y ≤ 6

Step-by-step explanation:

<u>Step 1: Convert words into an expression</u>

Forty six times y is no more than 276

46 * y ≤ 276

<u>Step 2: Solve for y</u>

46 * y / 46 ≤ 276 / 46

y ≤ 6

Answer: y ≤ 6

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3 years ago

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Resolve into partial fractions 7-5x/2x^2+x-1

Flauer [41]

The decomposition of partial fractions is to start with the simplified reply and then take it apart, to "decompose" the final expression into its initial polynomial fractions.

Given:

$\to \bold{\frac{7-5x}{2x^2+x-1}}\\\\$

To find:

partial fractions=?

Solution:

$\to \bold{\frac{7-5x}{2x^2+x-1}}\\\\\to \bold{\frac{7-5x}{2x^2+x(2-1)-1}}\\\\\to \bold{\frac{7-5x}{2x^2+2x-x-1}}\\\\\to \bold{\frac{7-5x}{2x(x+1)-1(x+1)}}\\\\\to \bold{\frac{7-5x}{(2x-1)(x+1)} = \frac{A}{(2x-1)} - \frac{B}{(x+1)} }\\\\\to \bold{7-5x = \frac{A ((2x-1)(x+1))}{(2x-1)} - \frac{B((2x-1)(x+1))}{(x+1)} }\\\\\to \bold{7-5x = A(x+1) -B(2x-1) }\\\\$

putting $x=-1$

$\to \bold{7-5(-1) = A(-1+1) -B(2(-1)-1) }\\\\\to \bold{7+5 = A(0) -B(-2-1) }\\\\\to \bold{12 = +3B }\\\\\to \bold{B = \frac{12}{3} }\\\\\to \bold{B = 4 }\\\\$

putting $x= \frac{1}{2}$

$\to \bold{7-5(\frac{1}{2}) = A(\frac{1}{2}+1) -B(2(\frac{1}{2})-1) }\\\\\to \bold{7-\frac{5}{2} = A(\frac{3}{2}) -B((\frac{2}{2})-1) }\\\\\to \bold{7-\frac{5}{2} = A(\frac{3}{2}) -B(1-1) }\\\\\to \bold{\frac{14-5}{2} = A(\frac{3}{2}) -B(0) }\\\\\to \bold{\frac{9}{2} = A(\frac{3}{2})}\\\\\to \bold{\frac{9}{2} \times \frac{2}{3} = A}\\\\\to \bold{\frac{9}{3} = A}\\\\\to \bold{A=3}\\\\$