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3 years ago

Solve the inequality and find the solution set. y-6 ≥12

1 answer:

4 0

The first step for solving this inequality is to move the constant to the right side by adding its opposites to both sides. This will look like the following:
y - 6 + 6 ≥ 12 + 6
Eliminate the opposites on the left side.
y ≥ 12 + 6
Lastly,, add the numbers on the right side together to get your final answer.
y ≥ 18
Let me know if you have any further questions.
:)

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the length of the sides of a square are initially 0 cm and increase at a constant rate of 11 cm per second. suppose the function

Likurg_2 [28]

The function of the area of the square is A(t)=121 $t^{2}$

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

$\frac{dx}{dt} =$ 11 $\frac{cm}{sec}$

⇒x=11t

Area of square= $(length of side)^{2}$

Area of square= $(11t)^{2}$ {as the length of side is 11t}{varies by time}

Area of square=121 $t^{2}$

Therefore,The function of the area of the square is A(t)=121 $t^{2}$

Learn more about The function of the area of the square is A(t)=121 $t^{2}$

Lets assume the length of side of square is x

$\frac{dx}{dt} =$ 11 $\frac{cm}{sec}$

⇒x=11t

Area of square= $(length of side)^{2}$

Area of square= $(11t)^{2}$ {as the length of side is 11t}{varies by time}

Area of square=121 $t^{2}$

Therefore,The function of the area of the square is A(t)=121 $t^{2}$

Learn more about area here:

brainly.com/question/27683633

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3 0

2 years ago

In the diagram below, ST is parallel to PQ. If ST is 4 more than PS,SR = 12, and PQ = 15, find the length of PS. Figures are not

Tpy6a [65]

Since segments ST and PQ are parallel, triangles SRT and PRQ are similar due to the AAA postulate. In general, the ratio between the corresponding sides of two similar triangles is constant; therefore,

$\frac{SR}{PR}=\frac{RT}{RQ}=\frac{TS}{QP}$

Furthermore,

$PS=PR-RS$

Finding PR and RS,

$\begin{gathered} ST=4+PS \\ SR=12 \\ PQ=15 \end{gathered}$

Then,

$\frac{12}{PR}=\frac{TS}{15}$ $\begin{gathered} \Rightarrow\frac{12}{PR}=\frac{4+PS}{15} \\ \Rightarrow\frac{12}{PS+12}=\frac{4+PS}{15} \end{gathered}$

Solving for PS,

$\begin{gathered} 12\cdot15=(PS+12)(PS+4) \\ \Rightarrow180=PS^2+16PS+48 \end{gathered}$

Solve the quadratic equation in terms of PS, as shown below

$\begin{gathered} \Rightarrow PS^2+16PS-132=0 \\ \Rightarrow PS=\frac{-16\pm\sqrt[]{16^2-4(-132)}}{2}=\frac{-16\pm28}{2} \\ \Rightarrow PS=-22,6 \end{gathered}$

And PS is a segment; therefore, it has to be positive.

Hence, the answer is PS=6

3 0

1 year ago

True or false and if false what would be the correct term be?

vagabundo [1.1K]

It’s True.. you take the “extreme” variable from each proportion and cross multiply them before setting them equal to the product of the two “means” (which are just the other 2 numbers in the proportions).

3 0

4 years ago

SOLVE.PLS HELP HURRY I PUT TEN POINTS! click to see full and the lines mean the estimate here is the question:estimate then add

dimaraw [331]

The first one is 17 13/15 the second one is 22 1/20

4 0

3 years ago

6( 5-8x) +12= -54

UkoKoshka [18]

<span>6( 5-8x) +12= -54
30-48x+12=-54 this comes from distributing(multiplying) 6x5 and 6 times -8x
30+12-48x=-54 you have to add the coefficients 30+12
42-48x=-54
-42 -42
-----------------------
-48x=-96 divide by -48
x=2</span>

5 0

3 years ago

Read 2 more answers