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3 years ago

6p-10c-8 when p=6 and c=8

Mathematics

1 answer:

ELEN [110]3 years ago

5 0

The answer is 36 in simplist form

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What is 123,499 rounded tot he nearest thousand

schepotkina [342]

125,000 because you would round it up

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3 years ago

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A student bought a book for $7.50 and a pen.the total cost was $9.50.which of the following can be used to find the cost of the

zloy xaker [14]

D: because if you take 9.50-7.50=2 you would get p=price or the price of the pen.
Hope I helped=)

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3 years ago

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Consider functions f and g.1 + 12f(1) = 12 + 4. – 12for * # 2 and 7 -64.2 – 16. + 1641 +48for a # -12 Which expression is equal

lisabon 2012 [21]

Given the following functions below,

$\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}\text{ and} \\ g(x)=\frac{4x^2-16x+16}{4x+48} \end{gathered}$

Factorising the denominators of both functions,

Factorising the denominator of f(x),

$\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}=\frac{x+12}{x^2+6x-2x-12}=\frac{x+12}{x(x+6)-2(x+6)}=\frac{x+12}{(x-2)(x+6)} \\ f(x)=\frac{x+12}{(x-2)(x+6)} \end{gathered}$

Factorising the denominator of g(x),

$\begin{gathered} g(x)=\frac{4x^2-16x+16}{4x+48}=\frac{4(x^2-4x+4)}{4(x+12)} \\ \text{Cancel out 4 from both numerator and denominator} \\ g(x)=\frac{x^2-4x+4}{x+12}=\frac{x^2-2x-2x+4}{x+12}=\frac{x(x-2)-2(x-2)}{x+12}=\frac{(x-2)^2}{x+12} \\ g(x)=\frac{(x-2)^2}{x+12} \end{gathered}$

Multiplying both functions,

$undefined$

4 0

1 year ago

Anyone help with this please? I need a, b and c

skad [1K]

I hope this helps you

7 0

3 years ago

Secants L J and L M intersect and form an angle at point L. Solve for x.

german

Answer:

x = 14

Step-by-step explanation:

Assume your diagram is like the one below.

The intersecting secant angles theorem states, "When two secants intersect outside a circle, the measure of the angle formed is one-half the difference between the far and the near arcs."

For your diagram, that means

$\begin{array}{rcl}m\angle L &=&\dfrac{1}{2} \left(m \widehat {JM} - m\widehat {PQ}\right)\\\\(3x + 13)^{\circ}& = &\dfrac{1}{2} \left[(8x + 48)^{\circ} - (5x - 20)^{\circ}\right]\\\\3x + 13& = &\dfrac{1}{2}(8x + 48 - 5x + 20)\\\\3x + 13& = &\dfrac{1}{2}(3x + 68)\\\\6x + 26 & = & 3x + 68\\6x & = & 3x + 42\\3x & = & 42\\x & = & \mathbf{14}\\\end{array}$

Check:

$\begin{array}{rcl}(3\times14 + 13) & = &\dfrac{1}{2} \left[(8\times14 + 48)^{\circ} - (5\times14 - 20)^{\circ}\right]\\\\42 + 13& = &\dfrac{1}{2}(112 + 48 - 70 + 20)\\\\55& = &\dfrac{1}{2}(110)\\\\55 & = & 55\\\end{array}$

It checks.

6 0

3 years ago