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Dimas [21]
3 years ago
15

Increase $80 by 10%....

Mathematics
2 answers:
nikklg [1K]3 years ago
8 0

Answer:

$88

Step-by-step explanation:

To find the increase of $80 by 10%, find 10% of $80 then add it by the original amount

10% of 80=80*0.1=$8

The original amount=$80

Total amount=$80+$8=$88

anygoal [31]3 years ago
5 0

Answer:88$

Step-by-step explanation:

10% of 80=80*0.1=$8

The original amount=$80

Total amount=$80+$8=$88

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The coordinates of the vertices of the image are L' = (-4, 6), M' = (-5, 1) and  N' = (-7, 3)

<h3>What are the coordinates of the vertices of the image?</h3>

The vertices of the preimage of the triangle are given as:

L = (4, -6)

M = (5, -1)

N = (7, -3)

The rotation is given as: 180 degrees counterclockwise

<h3 />

The rule of this rotation is

(x, y) => (-x, -y)

So, we have:

L' = (-4, 6)

M' = (-5, 1)

N' = (-7, 3)

Hence, the coordinates of the vertices of the image are L' = (-4, 6), M' = (-5, 1) and  N' = (-7, 3)

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1 year ago
I NEED THE SLOPE INTERCEPT FORM!!! ITS DUE BY 12
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Answer: y = 4x -1

Step-by-step explanation:

Given by the graph.

Y axis

and rise over run.

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2 years ago
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PolarNik [594]

Answer:

It is the y=intercept

Step-by-step explanation:

Because if there is no x-value and there is a Y-value,  y-intercept

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3 years ago
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A taxi charges a flat rate of $1.75 plus an additional $0.65 per mile. If Erica has at most $10 to spend on the cab ride, how fa
tekilochka [14]
1.75 + 0.65m < = 10
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8 0
3 years ago
Find a polynomial with integer coefficients that satisfies the given conditions. R has degree 4 and zeros 3 − 3i and 2, with 2 a
dolphi86 [110]

Answer:

The required polynomial is P(x)=x^4-10x^3+46x^2-96x+72.

Step-by-step explanation:

If a polynomial has degree n and c_1,c_2,...,c_n are zeroes of the polynomial, then the polynomial is defined as

P(x)=a(x-c_1)(x-c_2)...(x-x_n)

It is given that the polynomial R has degree 4 and zeros 3 − 3i and 2. The multiplicity of zero 2 is 2.

According to complex conjugate theorem, if a+ib is zero of a polynomial, then its conjugate a-ib is also a zero of that polynomial.

Since 3-3i is zero, therefore 3+3i is also a zero.

Total zeroes of the polynomial are 4, i.e., 3-3i, 3_3i, 2,2. Let a=1, So, the required polynomial is

R(x)=(x-3+3i)(x-3-3i)(x-2)(x-2)

R(x)=((x-3)+3i)((x-3)-3i)(x-2)^2

R(x)=(x-3)^2-(3i)^2((x-3)-3i)(x-2)^2     [a^2-b^2=(a-b)(a+b)]

R(x)=(x^2-6x+9-9(i)^2((x-3)-3i)(x-2)^2

R(x)=(x^2-6x+18)(x^2-4x+4)                [i^2=-1]

R(x)=(x^2-6x+18)(x^2-4x+4)

R(x)=x^4-10x^3+46x^2-96x+72

Therefore the required polynomial is P(x)=x^4-10x^3+46x^2-96x+72.

3 0
3 years ago
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