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3 years ago

What is the Y-intercept ??

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

6 0

The Y intercept is (0,2) so you were correct in your answer

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Thank you guys for helping me I know I'm being annoying

Ierofanga [76]

A² + b² = c²

c² = 4² + 6.5²

c² = 58.25

c = 7.63

Perimeter = 4 + 6.5 + 7.63 = 18.13 units.

Perimeter after the increase in scale = 18.13 x 4 = 72.52 units.

Answer: 72.52 units

8 0

2 years ago

One year ago, Joan was five times as old as her pony. In one year's time, the sum of the ages will be 22. How old is Joan now?

iogann1982 [59]

Answer:

22 divided 5 I would think

Step-by-step explanation:

4 0

2 years ago

Prove that:<br><br>cos20°cos40°cos80°=1/8

Naya [18.7K]

Answer:

see explanation

Step-by-step explanation:

Using the double angle identity for sine

sin2x = 2sinxcosx

Consider left side

cos20°cos40°cos80°

= $\frac{1}{2sin20}$ (2sin20°cos20°)cos40°cos80°

= $\frac{1}{4sin20}$ (2sin40°cos40°)cos80°

= $\frac{1}{4sin20}$ (sin80°cos80° )

= $\frac{1}{8sin20}$ (2sin80°cos80° )

= $\frac{1}{8sin20}$ . sin160°

= $\frac{1}{8sin20}$ . sin(180 - 20)°

= $\frac{1}{8sin20}$ . sin20°

= $\frac{1}{8}$ = right side , thus proven

7 0

2 years ago

Read 2 more answers

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago

The area of this parallelogram is 120 ft. Find the value of h

julsineya [31]

are you going to show a picture??? its impossible to answer otherwize

4 0

3 years ago