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sasho [114]
3 years ago
5

Solve for y 5x-2v=15*please help!!​

Mathematics
1 answer:
lakkis [162]3 years ago
7 0

Answer: -2 + 5x - 15 = 0

Step-by-step explanation: you have to subtract 5x and 2v so i had to switch them around for a proper answer then subtract 15 from -2v and 5x and you get 0

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Which is the distance from point A to B?<br><br> A) 3<br><br> B) 4<br><br> C) 5<br><br> D) 6
Andre45 [30]

Answer:

c) 5

Step-by-step explanation:

In this question you are being asked to find the hypotenuse of the triangle

the triangle has one side that is 3 and the other is 4

use Pythagoras to find the missing side (a^2+b^2=c^2)

3^2+4^2=c^2

simplify the exponents

9+16=c^2

combine like terms/adding

25=c^2

take the square root

5=c

the length of the hypotenuse is 5

4 0
3 years ago
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Help me to do this math question ​
viva [34]

Answer:

b) q

c) 7a

d) 2y

Step-by-step explanation:

b) q must be added to p to get p+q

c) 12a-7a = 5a

d) 2y is subtracted from 3x

5 0
3 years ago
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Triangle ABC contains the point A(5,2). Find the coordinates of A' after a reflection over the x-axis
Sholpan [36]

Answer:

Step-by-step explanation:

Should be 5,-2 OR NOT

4 0
4 years ago
What is equivalent to 16.8 - 18.6
In-s [12.5K]

Answer:

16.8-18.6= -1.8

Step-by-step explanation:

It must be negative

4 0
3 years ago
For p = $3000, r = 3.5%,and t = 5 years, Find the balance in an account When interest is compounded (a) quarterly, (b) monthly,a
MAVERICK [17]

Answer:

a) $3,571.02

b) $3,572.9

c) $3,573.74

Step-by-step explanation:

Data provided in the question:

p = $3000,

r = 3.5%,

t = 5 years

a) quarterly

number of periods in a year, n = 4

Interest rate per period = 3.5% ÷ 4 = 0.875%

Now,

A = p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}

A = total amount

n = number of times compounded per year

on substituting the respective values, we get

A = 3000 × \left( 1 + \frac{ 0.035 }{ 4 } \right)^{\Large{ 4 \cdot 5 }}

A = 3000 × [/tex]\cdot { 1.00875 } ^ { 20 }[/tex]

A = 3000 × 1.19034

A = $3,571.02

b) monthly

number of periods in a year, n = 12

Now,

A = p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}

on substituting the respective values, we get

A = 3000 × \left( 1 + \frac{ 0.035 }{ 12 } \right)^{\Large{ 12 \cdot 5 }}

A = 3000 × [/tex]\cdot { 1.002917} ^ { 60 }[/tex]

A = 3000 × 1.190967

A = $3,572.9

c) continuously

A = pe^{r\times t}

on substituting the respective values, we get

A = 3,000 × e^{0.035\times 5}

or

A = 3,000 × e^{0.175}

or

A = 3,000 × 1.1912

or

A = $3,573.74

5 0
3 years ago
Read 2 more answers
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