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2 years ago

Geometry math problem

Mathematics

2 answers:

malfutka [58]2 years ago

6 0

Answer:

uhhhhh

Step-by-step explanation:

Anastaziya [24]2 years ago

4 0

Answer:

49 degrees???;!!!!!!

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What is the square root of 38 to the nearest tenth

Ipatiy [6.2K]

6.16441 which rounds to 6.2 in the tenths place

5 0

2 years ago

If s(x)=x-7 and t(x) =4x^2+3 which expression is equivalent to (t×s)(x)

marshall27 [118]

Answer:

$\large\boxed{(t\times s)(x)=4x^3-28x^2+3x-21}$

Step-by-step explanation:

$s(x)=x-7\\t(x)=4x^2+3\\\\(t\times s)(x)=t(x)\times s(x)\\\\\text{Substitute:}\\\\(t\times s)(x)=(4x^2+3)(x-7)\\\\\text{use FOIL:}\ (a+b)(c+d)=ac+ad+bc+bd\\\\(t\times s)(x)=(4x^2)(x)+(4x^2)(-7)+(3)(x)+(3)(-7)\\\\=4x^3-28x^2+3x-21$

8 0

3 years ago

Which is the completely factored form of 12x3 – 60x2 + 4x – 20?

Natalija [7]

Answer:

$4(x-5)(3{x}^{2}+1)$

Step-by-step explanation:

1) Find the Greatest Common Factor (GCF).

1 - What is the largest number that divides evenly into $12x^3,-60x^2,4x,$ and $-20$ ?

It is $4.$

2 - What is the highest degree of $x$ that divides evenly into $12x^3,-60x^2,4x,$ and $-20$ ?

It is 1, since $x$ is not in every term.

3 - Multiplying the results above,

The GCF is 4.

2) Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)

$4(\frac{12{x}^{3}}{4}+\frac{-60{x}^{2}}{4}+\frac{4x}{4}-\frac{20}{4})$

3) Simplify each term in parentheses.

$4(3{x}^{3}-15{x}^{2}+x-5)$

4) Factor out common terms in the first two terms, then in the last two terms.

$4(3{x}^{2}(x-5)+(x-5))$

5) Factor out the common term $x-5$ .

$4(x-5)(3{x}^{2}+1)$

5 0

1 year ago

eight less than four times a number is twelve. what is the number? the options are, 5 , 1 , -3 , -5 , please i need a answer asa

ra1l [238]

4x minus 8 = 12 answer: 5

6 0

3 years ago

Use Cramer's rule to find the solution to the following system of linear equations.

slavikrds [6]

Answer:

x = 53

y = 30

Step-by-step explanation:

Step(I):-

Given equations are

x -2y =-7 ...(I)

5x-9y =-5 ..(ii)

The matrix form AX = B

$\left[\begin{array}{ccc}1&-2\\ 5 & -9\\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}-7\\-5\\\end{array}\right]$