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4 years ago

531 = ( X × 6) + 21 LAST QUESTIONNNN

2 answers:

4 0

To solve for X: First you multiply x and 6 then subtract 21 on both sides and then divide by 6

531 = ( X × 6) + 21
531 = 6X + 21
510 = 6X
85 = X

Hope this helps :)

Karo-lina-s [1.5K]4 years ago

4 0

First, subtract 21 from 531, which should get you 510. Then divide that by 6 which is 85. Then plug 85 in for x and solve to see if you get 531 :)

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Guys plzzz help me i don't know how to figure out this question 10(6x+7)=26

Mekhanik [1.2K]

What is x
First you use distributive property
60x+70=26
Next subtract 70 on both sides which gives you -44
60x=-44
Then you divide 60 into -44 which will give you -1.36__36 is repeathing
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7 0

3 years ago

Simplify. x/4x+x^2

djyliett [7]

$\bf \cfrac{x}{4x+x^2}\implies \cfrac{\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}{\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(4+x)}\implies \cfrac{1}{4+x}\qquad \{x|x\in \mathbb{R}, x\ne -4\}$

if you're wondering about the restriction of x ≠ -4, is due to that would make the fraction with a denominator of 0 and thus undefined.

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4 years ago

Sam's Fruit Cellar earns a $0.40 profit for each apple that it sells and a $0.40 profit for each banana that it sells. Which exp

Vera_Pavlovna [14]

Answer:

Total Profit = P(a,b) = ($0.40a) + ($0.40b), where a = # of apples sold and b = # of bananas sold.

6 0

3 years ago

Read 2 more answers

Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

Question 1 a. i) Find the value of x.

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

Question 1a ii) Find the volume of the frustrum

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

Question 1b. Calculate the volume of the bin

i) Upper frustrum

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

ii) Lower frustrum

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

iii) Add the volume of the two frustrums

$Volume=27.6m^3+38.3m^3=65.9m^3$

6 0

3 years ago

Which of the following ratios is used in finding the tangent of an angle in a right triangle?

Softa [21]

Looking at the triangle from the aspect of the reference angle, the tangent of that angle would be the ratio of the side opposite the angle over the side adjacent to the angle (which is NOT the hypotenuse!)

8 0

3 years ago