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IRINA_888 [86]
3 years ago
9

4{[5 (x-3)+2]-3 [2 (x+5)-9]}

Mathematics
2 answers:
lapo4ka [179]3 years ago
8 0
<span>Solve by applying PEMDAS - the order of operations:
P = Parenthesis or brackets 
E = Exponents or radicals whichever comes first
M = Multiplication or division whichever comes first
D = Division or multiplication whichever comes first
A = Addition or subtraction whichever comes first
S = Subtraction or addition whichever comes first
.
4{[5(x-3)+2]-3[2(2x+5)-9]

4{[5x-15+2]-3[4x+10-9]}
4{[5x-13]-3[4x+1]}
4{[5x-13]-(12x+3)}
4{[5x-13]-12x+3}
4{5x-13-12x+3}
4{-7x-10}
-28x-40 
</span>
Over [174]3 years ago
6 0
Pemdas
parenthasees exponents mult/division additon/subtraction

parethenasees
x-3 and x+5 we can't do anything with so next
5(x-3)=5x-15

2(x+5)=2x+10

5x-15+2=5x-13

(2x+10-9)=2x+1

(5x-13)-3(2x+1)
-3(2x+1)=-6x-3

5x-13-6x-3=-x-16

4(-x-16)=-4x-64

the answer is -4x-64
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andreyandreev [35.5K]

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

We have

(r,\ 6),\ (5,\ -4)\\\\m=\dfrac{5}{7}

Substitute

\dfrac{-4-6}{5-r}=\dfrac{5}{7}\\\\\dfrac{-10}{5-r}=\dfrac{5}{7}\qquad|\text{cross multiply}\\\\5(5-r)=(-10)(7)\qquad|\text{use distributive property}\\\\25-5r=-70\qquad|-25\\\\-5r=-95\qquad|:(-5)\\\\\boxed{r=19}

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3 years ago
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djyliett [7]

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8 0
3 years ago
Please someone help me to prove this. ​
dem82 [27]
<h3><u>Answer</u> :</h3>

We know that,

\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}

\dag\bf\:sin2A=2sinA\:cosA

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

<u>Now, Let's solve</u> !

\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}

\leadsto\:\sf\dfrac{\frac{1-cos2A}{2}-\frac{1-cos2B}{2}}{\frac{2sinA\:cosA}{2}-\frac{2sinB\:cosB}{2}}

\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}

\leadsto\:\sf\dfrac{2sin\frac{2A+2B}{2}\:sin\frac{2A-2B}{2}}{2sin\frac{2A-2B}{2}\:cos\frac{2A+2B}{2}}

\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}

\leadsto\:\bf{tan(A+B)}

5 0
3 years ago
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Christine has $875.83 In her savings account the account pays 9% compounded monthly. Christine does not make any deposits or wit
Karolina [17]

Answer: the account earns interest of $40.16

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 875.83

r = 9% = 9/100 = 0.09

n = 12 because it was compounded 12 times in a year.

t = 6 months = 6/12 = 0.5 year

Therefore,.

A = 875.83(1+0.09/12)^0.5 × 12

A = 875.83(1+0.0075)^6

A = 875.83(1.0075)^6

A = 915.99

The interest that she earns is

915.99 - 875.83 = $40.16

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3 years ago
Please answer this ASAP! (No explanation needed!) &lt;3
stepan [7]

Answer:

Step-by-step explanation:

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