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4 years ago

Help please? thanks :)

Mathematics

1 answer:

DaniilM [7]4 years ago

6 0

The answer is C, this is bc that's the highest point on the parabola.

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One of the angles is 26°. The second angle is complementary to the first angle. What is the measure of the third angle?

lubasha [3.4K]

2nd is 64 3rd is 90 all 3 = 180

6 0

4 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

What is the height of a right triangle with an angle that measures 30 degrees adjacent to a base of 14

stepladder [879]

The height of the right triangle is 8.08

<h3>Calculating the height of a triangle </h3>

From the question, we are to determine the height of the described right triangle

From the given information,

The angle measure is 30 degrees adjacent to the base

and

The base is 14

Using SOH CAH TOA

Adjacent = 14

Opposite = height of the triangle

Let the height the h

∴ Opposite = h

Thus,

tan 30° = h/14

h = 14 × tan 30°

h = 8.08

Hence, the height of the right triangle is 8.08

Learn more on Calculating height of triangle here: brainly.com/question/10082088

#SPJ1

3 0

1 year ago

Pls help! will give brainlist!

lara31 [8.8K]

It couldn’t be anything else but C line reflection i’m assuming

4 0

3 years ago

Read 2 more answers

Translate the phrase twice m increased by 2 is equal to 7 more then m

yaroslaw [1]

The answer to this question is B

3 0

3 years ago