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Serjik [45]
3 years ago
14

X over 13 is greater than or equal to 4... answer?

Mathematics
2 answers:
Olin [163]3 years ago
5 0
 x   ≥  4   is x/13 times 13 ≥ 4 times 13
13

That means that x ≥ 52.

BaLLatris [955]3 years ago
4 0
In mathematical expression, You can write it as:
x/13 ≥ 4

Multiply both sides by 13, 
x/13 * 13 ≥ 4 * 13
x ≥ 52

In short, Your Answer would be x ≥ 52

Hope this helps!
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If f(x) = -2x – 14, then f-(x) =
jok3333 [9.3K]

Answer:

Is it f(x)=2x+14?

Step-by-step explanation:

8 0
3 years ago
Solve 2x2 − 12x + 20 = 0.<br><br> 3 ± i<br> 3 ± 2i<br> 1 ± 2i<br> 2 ± 3i
makkiz [27]
This is the concept of quadratic equations, solving the equation 2x^2-12x+20=0 using the formula we get:
x=[-b+/-sqrt(b^2-4ac)]/(2a)
from our equation, a=2,b=-12 and c=20
x=[-(-12+/-sqrt((-12)^2-4*2*20)]/(2*2)
x=[12+/-sqrt(144-160]/4
x=[12+/-sqrt(-16)]/4
x=[12+/-4i]/4
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the answer is A] 3+/-i
7 0
3 years ago
A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
mr Goodwill [35]

Answer: 8.26 m

Step-by-step explanation:

$$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$$

$$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$$

$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}

$$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$$

$$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$$

$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square

8 0
2 years ago
35°<br> X=<br> solve for the missing angles
Vedmedyk [2.9K]
Heheh I am 103 for this
8 0
3 years ago
What are the solutions to the equation x2−12x+100=0?
marysya [2.9K]

Answer:

x=6+8i

x=6-8i

Step-by-step explanation:

x^2-12x+100=0

solve by quadratic formula.

\frac{-b+-\sqrt{b^2-4(a*c)} }{2*a}

\frac{12+\sqrt{(-12)^2-4*(1*100)} }{2*1}

\frac{12-\sqrt{(-12)^2-4*(1*100)} }{2*1}

which is

x=\frac{12+16i}{2}

x=\frac{12-16i}{2}

Simplify.

x=6+8i

x=6-8i

Solved.


Hope this helps.

8 0
2 years ago
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