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3 years ago

In the graph 5x2 + 3, the y-intercept is....

Mathematics

1 answer:

ira [324]3 years ago

5 0

Answer:

0 zero

Step-by-step explanation:

The slope-intercept form is y=mx+b y = m x + b , where m m is the slope and b b is the y-intercept. Combine x x and 13 1 3 . Rewrite in slope-intercept form. Using the slope-intercept form, the y-intercept is 0 .

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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

Solve for x and justify your answer

Scorpion4ik [409]

The answer is 51°.

The top two lines are parallel to each other. Therefore alternate interior angles theorem applies. The bottom right corner that is underneath the line (for lack of a better description) is also 58°.

We also know that a triangle equals 180°. So 58+71= 129°.

So to find that missing angle we subtract 129 from 180 to get the 51°.

3 0

3 years ago

i need help finding the area of a figure where the rectnagles area is 5 x 14 and the half circles diameter is 6

masha68 [24]

Answer:

84.14 square units, total

Step-by-step explanation:

Use the formulas for area of the figures you're interested in.

<u>Rectangle Area</u>

A = LW

<u>Circle Area</u>

A = πr^2

Your rectangle area will be ...

A = 5 × 14 = 70 . . . . square units

The area of a full circle with diameter 6 (radius = 6/2 = 3) will be ...

A = π × 3^2 = 9π

The area of a semicircle of that size is half that:

semicircle area = 9π/2 = 4.5π

Then the total area of the two figures is ...

70 + 4.5π ≈ 70 +14.14 = 84.14 . . . square units

8 0

3 years ago

If kayleigh is 102 centimeter tall is she greater than or less than 1 meter

Illusion [34]

Since there are 100 centimeters in 1 meter, Kayleigh is greater than 1 meter.

7 0

3 years ago

Read 2 more answers

PLZ HELP ME ON THIS ONLY PROBLEM???

Rom4ik [11]

Exact Form:
9
-
16

Decimal Form:
0.5625
That should be it ! :)

8 0

3 years ago

Read 2 more answers