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3 years ago

Please help thanks :)

Mathematics

2 answers:

Aleksandr-060686 [28]3 years ago

6 0

Answer: 1 song is 1.3 the total cost of 25 songs would be 33.75

strojnjashka [21]3 years ago

5 0

Answer:

1 song is 1.3 the total cost of 25 songs would be 33.75.

Step-by-step explanation:

have a nice life

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Find the value of the combination. 5C5 A:1 B:5 C:0

Olenka [21]

Answer:

A:1

Step by step explanations:

C(n,r)=C(5,5)

n=5

r=5

=n! / (r!(n-r)!)

=5! / (5!(5-5)!)

6 0

3 years ago

What is the range of the function shown on the graph above?

OverLord2011 [107]

Answer:

-9 ≤ y ≤ 8

Step-by-step explanation:

look at the endpoints of the segment; (7, 8) and (-2, -9)

the y-values represent the range and they go from -9 to 8, inclusive

8 0

3 years ago

Ryan bought a sofa on sale for $342. This price was 28% less than the original price.

Snezhnost [94]

The correct answer is $475

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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

A sailboat leaves port and sails 12 kilometers west and then 9 kilometers north. The sailboat is now kilometers from port

Luden [163]

Answer: The sailboat is at a distance of 15 km from the port.

Step-by-step explanation: Given that a sail boat leaves port and sails 12 kilometers west and then 9 kilometers north.

We are to find the distance between the sailboat from the port in kilometers.

Since the directions west and north are at right-angles, we can visualize the movement of the sailboat in the form of a right-angled triangle as shown in the attached figure.

The sailboat moves leaves the port at P and reach O after sailing 12 km west. From point O, again it moves towards north 9 km and reach the point S.

PS = ?

Using the Pythagoras theorem, we have from right-angled triangle SOP,

$PS^2=OS^2+OP^2=9^2+12^2=81+144=225\\\\\Rightarrow PS=15~\textup{km}.$

Thus, the sailboat is at a distance of 15 km from the port.

5 0

4 years ago