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3 years ago

2x - 5 < 33 what are all the values to this inequality

Mathematics

1 answer:

Lynna [10]3 years ago

4 0

Answer:

x < 19

Step-by-step explanation:

Given that,

An inequality 2x - 5 < 33.

LHS of the inequality is 2x - 5

RHS of the inequality is 33

LHS < RHS

i.e.

2x - 5 < 33

Add 5 to both sides of the inequality

2x - 5 +5 < 33+5

2x < 38

Divide both sides by 2.

x < 19

So, the value of x is less than 19.

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If the point (3,4) is 6 units above the point P(x,y) and 11 units to the right of it, then what are the coordinates of the point

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Answer:

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The total cost, in dollars, to order x units of a certain product is modeled by C(x)=5x²+320 . According to the model, for what

lora16 [44]

Answer:

The minimum cost per unit is obtained for an order of 8 units.

Step-by-step explanation:

Since the total cost is modeled by;

C(x)=5x²+320

Then;

1 unit costs; C(x)=5(1)²+320 = 325

cost per unit 325/1 = 325

8 units costs; C(x)=5(8)²+320 = 640

cost per unit = 640/8 = 80

80 units costs; C(x)=5(80)²+320 = 32320

Cost per unit = 32320/80 = 404

The minimum cost per unit is obtained for an order of 8 units.

7 0

3 years ago

How can 7 1/5+(-8 3/5) be expressed as the sum of its integer and fractional parts

Nadya [2.5K]

Answer:

see the explanation

Step-by-step explanation:

we know that

A mixed number is the sum of a integer and a fractional part

we have

$7\frac{1}{5}+(-8\frac{3}{5})$

$7\frac{1}{5}=7+\frac{1}{5}$

$-8\frac{3}{5}=-8-\frac{3}{5}$

substitute

$7+\frac{1}{5}+(-8-\frac{3}{5})$

Group terms

$(7-8)+(\frac{1}{5}-\frac{3}{5})$

$(-1)+(-\frac{2}{5})$

$-1\frac{2}{5}$

6 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

2x - 5 < 33 what are all the values to this inequality​

2x - 5 < 33 what are all the values to this inequality