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kherson [118]
3 years ago
11

The pizza costs £18.00. Mark agrees to pay 1/4 of the cost. Samuel agrees to pay 1/2 of the cost Imran agrees to pay the rest. H

ow much does Imran pay?
Mathematics
1 answer:
MariettaO [177]3 years ago
3 0

Answer:

4.5 or 4.50 dollars

Step-by-step explanation:

if we split 18.00 into \frac{1}{4} each part of the fraction would be 4.5 . or 4.50

mark pays 4.50

samuel pays 9.00

add those both and you'll get 13.50 $

now subtract 13.50 from 18.00 ( og price)

imran pays 4 . 5 0

hope this helped :p

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To test the effectiveness of a business school preparation course, 8 students took a general business test before and after the
Lisa [10]

Solution :

Group   Before     After

Mean    693.75    743.75

Sd         155.37     143.92

SEM        54.93     50.88

n              8            8

Null hypothesis : The preparation course not effective.

$H_0: \mu_d = 0$

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4 0
2 years ago
Prove the divisibility:<br><br>45^45·15^15 by 75^30
garri49 [273]

Answer:

3^{75}.

Step-by-step explanation:

We have been an division problem: \frac{45^{45}*15^{15}}{75^{30}}.

We will simplify our division problem using rules of exponents.

Using product rule of exponents (a*b)^n=a^n*b^n we can write:

45^{45}=(9*5)^{45}=9^{45}*5^{45}

15^{15}=(3*5)^{15}=3^{15}*5^{15}

75^{30}=(15*5)^{30}=15^{30}*5^{30}

Substituting these values in our division problem we will get,

\frac{9^{45}*5^{45}*3^{15}*5^{15}}{15^{30}*5^{30}}

Using power rule of exponents a^n*a^m=a^{n+m} we will get,

\frac{9^{45}*5^{(45+15)}*3^{15}}{15^{30}*5^{30}}

\frac{9^{45}*5^{60}*3^{15}}{15^{30}*5^{30}}

Using product rule of exponents (a*b)^n=a^n*b^n we will get,

\frac{(3*3)^{45}*5^{60}*3^{15}}{(3*5)^{30}*5^{30}}

\frac{3^{45}*3^{45}*5^{60}*3^{15}}{3^{30}*5^{30}*5^{30}}

Using power rule of exponents a^n*a^m=a^{n+m} we will get,

\frac{3^{(45+45+15)}*5^{60}}{3^{30}*5^{(30+30)}}

\frac{3^{105}*5^{60}}{3^{30}*5^{60}}

\frac{3^{105}}{3^{30}}

Using quotient rule of exponent \frac{a^m}{a^n}=a^{m-n} we will get,

\frac{3^{105}}{3^{30}}=3^{105-30}

3^{105-30}=3^{75}

Therefore, our resulting quotient will be 3^{75}.

7 0
3 years ago
Find the constant $a$ such that\[(x^2 - 3x + 4)(2x^2 +ax + 7) = 2x^4 -11x^3 +30x^2 -41x +28.\]
Colt1911 [192]

The value of constant a is -5

Further explanation:

We will use the comparison of co-efficient method for finding the value of a

So,

Given

(x^2 - 3x + 4)(2x^2 +ax + 7)\\= x^2((2x^2 +ax + 7)-3x((2x^2 +ax + 7)+4(2x^2 +ax + 7)\\= 2x^4+ax^3+7x^2-6x^3-3ax^2-21x+8x^2+4ax+28\\Combining\ alike\ terms\\=2x^4+ax^3-6x^3+7x^2-3ax^2+8x^2-21x+4ax+28\\= 2x^4 +(a-6)x^3+(15-3a)x^2-(21-4a)x+28\\

As it is given that

(x^2 - 3x + 4)(2x^2 +ax + 7) = 2x^4 -11x^3 +30x^2 -41x +28

In this case, co-efficient of variables will be equal, so we can compare the coefficients of x^3, x^2 or x

Comparing coefficient of x^3

a-6 = -11\\a = -11+6\\a = -5

So the value of constant a is -5

Keywords: Polynomials, factorization

Learn more about factorization at:

  • brainly.com/question/1414350
  • brainly.com/question/1430645

#LearnwithBrainly

4 0
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