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bija089 [108]
3 years ago
5

Given thatf(x)=x5−x,findf(−15).

Mathematics
1 answer:
bekas [8.4K]3 years ago
6 0
The answer is 60. You plus 15 into x
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Use the grouping method to factor the polynomial below completely. x3 - 3x2 + 5x - 15
LUCKY_DIMON [66]

Answer:

(x²+5) (x-3)²

Step-by-step explanation:

x³ - 3x² + 5x - 15

x²(x-3) + 5(x-3)

(x²+5) (x-3)²

4 0
3 years ago
Write the point-slope form of the equation of the line through the given point with the given
UNO [17]

Answer:

The point-slope form of the equation is:

  • y-\left(-5\right)=-3\left(x-3\right)

Step-by-step explanation:

Given

  • point (3, -5)
  • slope = -3

We know the point-slope form of the line equation is

y-y_1=m\left(x-x_1\right)

where m is the slope

substituting the values m = -3 and the point (3, -5)

y-y_1=m\left(x-x_1\right)

y-\left(-5\right)=-3\left(x-3\right)

Thus, the point-slope form of the equation is:

  • y-\left(-5\right)=-3\left(x-3\right)
6 0
3 years ago
Please someone what is the total surface area of this pryamid ​
Dimas [21]

Answer:

\large\boxed{C.\ (25+25\sqrt3)\ in^2}

Step-by-step explanation:

We have the square and four equilateral triangles.

The formula of an area of a squre:

A_S=a^2

a - length of side

The formula of an area of an equilateral triangle:

A_T=\dfrac{a^2\sqrt3}{4}

a - length of side

Clculate the areas:

SQURE:

A_S=x^2\ in^2

TRIANGLE:

A_T=\dfrac{x^2\sqrt3}{4}\ in^2

The SURFACE AREA of a square pyramid:

S.A.=A_S+4A_T\\\\S.A.=x^2+4\cdot\dfrac{x^2\sqrt3}{4}=x^2+x^2\sqrt3

Put x = 5:

S.A.=5^2+5^2\sqrt3=(25+25\sqrt3)\ in^2

7 0
3 years ago
Please help and show work!!
Jlenok [28]

A(1, 1), B(7, 1), C(1, 9)

AB = 7 - 1 = 6

AC = 9 - 1 = 8

Use the Pythagorean theorem:

AB² + AC² = BC²

Substitute:

BC² = 6² + 8²

BC² = 36 + 64

BC² = 100 → BC = √100 → BC = 10

The perimeter of ΔABC:

P = 6 + 8 + 10 = 24

<h3>Answer: 24 units</h3>
5 0
3 years ago
Find two consecutive whole numbers that √45 falls between.
Hoochie [10]
It would fall between 6 and 7.
7 0
3 years ago
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