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4 years ago

What Is 7 subtracted from a number if “A Number” is x

Mathematics

1 answer:

julia-pushkina [17]4 years ago

3 0

X-7 is the answer because youre taking seven from a variable

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Can 22 w-(-14 w) be simplified?

olga nikolaevna [1]

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3 years ago

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How many lines of symmetry does the infield of a baseball field have?

Gnesinka [82]

The inside of the feild has 4 lines if symmetry

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4 years ago

Indiras spring garden has half as many tulips as daffodils and three times as many hyacinths as tulips. If there are a total of

kow [346]

Indira is having 34 tulips, 68 daffodils and 102 hyacinths , altogether 204 flowers in her spring garden.

Given, In Indira's springtime garden, there are three times more hyacinths than tulips and half as many daffodils as tulips.

let the number of tulips be = x

the number of daffodils be = y

the number of hyacinths be = z

hence according to the question,

tulips are (x) = 1/2 (y) = y/2

hyacinths are (z) = 3(x) = 3x

therefore, total flowers = 204

tulips ₊ daffodils ₊ hyacinths = 204

y/2 ₊ y ₊ 3x = 204

substitute x value in the above equation.

y/2 ₊ y ₊ 3(y/2) = 204

take LCM

y ₊ 2y ₊ 3y = 408

6y = 408

y = 408/6

y = 68

hence the number of daffodils is 68.

now substitute y value in x = y/2

x = 68/2

x = 34

now substitute x value in z= 3x

z = 3(34)

z = 102

hence we get the number of tulips,daffodils and hyacinths as 34 , 68 and 102 respectively.

Learn more about Solving equations here:

brainly.com/question/13729904

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6 0

1 year 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$

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2 years ago

which of tthe following fucngions describes the average expense fir a player played, in terms of x, the number of games played?

Firdavs [7]

And we are waiting to see the functions listed?

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4 years ago