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3 years ago

HELP HELP HELP HELP HELP

Mathematics

2 answers:

slega [8]3 years ago

7 0

Answer:

a: $-20+12y$

Step-by-step explanation:

$10-6\left(5-2y\right)$

Expand:

$10-6* \:5-\left(-6\right)* \:2y$

$10-6* \:5+6* \:2y$

$10-30+12y$

Subtract:

$-20+12y$

mestny [16]3 years ago

5 0

Your question has been heard loud and clear.

10-6(5-2y) = 10 - 30 +12y = -20+12y.

(All you have to do is multiply -6 with 5 and -2y , then simplify the equation)

Thank you

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Does anyone know to solve this

Alekssandra [29.7K]

Answer:

slope =x2-x1/y2-y1

= 0-6/4-0

=-6/4

8 0

2 years ago

What is the additive inverse of th polynomial -9xy2+6x2y-5x3?

Norma-Jean [14]

Additive inverse of 9xy² + 6x²y - 5x³

-1(9xy² + 6x²y - 5x³) ⇒ (-1)(9xy²) + (-1)(6x²y) + (-1)(-5x³)

<span>-9xy² - 6x²y + 5x³ is the additive inverse of the given polynomial.</span>

5 0

3 years ago

Suppose that $1200 is invested at 612%, compounded quarterly. How much is in the account at the end of 5 years?

asambeis [7]

Answer:

$\$1,656.50$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=5\ years\\ P=\$1,200\\ r=6\frac{1}{2}\%=6.5\%=6.5/100=0.065\\n=4$

substitute in the formula above

$A=1,200(1+\frac{0.065}{4})^{4*5}$

$A=1,200(1.01625)^{20}$

$A=\$1,656.50$

5 0

3 years ago

Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²

beks73 [17]

<h2>Answer:</h2>

The derivative of the function f(x) is:

$f'(x)=-2x$

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as:

$f(x)=5-x^2$

We have:

$f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)$

( Since,

$(a+b)^2=a^2+b^2+2ab$ )

Hence, we get:

$f(x+h)=5-x^2-h^2-2xh$

Also, by using the definition of f'(x) i.e.

$f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Hence, on putting the value in the formula:

$f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x$

Hence, the derivative of the function f(x) is:

$f'(x)=-2x$

3 0

3 years ago

Read 2 more answers

Which of the following statements best describes the "grouping method" of factoring trinomials?

Masja [62]

Answer:

<em>B. The grouping method of factoring trinomials involves rewriting the bx term into the factors that fit the particular trinomial, and factoring these four terms using grouping</em>

Explanation:

The description may be better explained by applying it to an example.

Example:

trinomial: x² - x - 30

the general form of a trinomial is a x² - bx - 30

comparing with x² - x - 30 the <em>bx term </em>is - x

then you must <em>rewrite the bx term, - x,</em> into two terms whose coefficients are factors of 30:

Two numbers which add up - 1 and multipled are - 30. Those numbers are - 6 and + 5, because -6 + 5 = - 1 and (-6) × (+5) = -30.

Hence, the two terms are -6x and 5x, and the expression rewritten is:

x² - 6x + 5x - 30

<em>factor these four terms using grouping</em>:

(x² - 6x) + (5x - 30)

x(x - 6) + 5(x - 6)

(x - 6) (x + 5)

Hence, the factored trinomial is (x - 6) (x + 5)

4 0

3 years ago