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Convert 3.5 ⋅ 104 to standard form. 35,000 3,500 0.00035 0.000035

Paraphin [41]

35,000
............................................

6 0

3 years ago

Read 2 more answers

Which statement is true

EastWind [94]

Answer:

a

Step-by-step explanation:

6 0

4 years ago

A parabola can be drawn given a focus of ( - 8,0) and a directrix of y= - 2 What can be said about the parabola?

Tcecarenko [31]

Answer:

The equation of the parabola is:

y

=

1

32

x

2

Explanation:

Please notice that we know that the parabola opens upward or downward, because the directrix is a horizontal line:

y

=

−

8

This tells use that the equation of the parabola of the form:

y

=

f

(

x

)

Not

x

=

f

(

y

)

.

Furthermore, we can see that it must open upward, because the y coordinate of the focus is above the directrix.

The vertex has the same x coordinate as the focus:

x

=

0

And its y coordinate is halfway between the focus and the directrix:

y

=

8

−

8

2

=

0

Therefore the vertex is

(

0

,

0

)

Use the vertex form of the equation of a parabola that opens upward or downward:

y

=

a

(

x

−

h

)

2

+

k

where the vertex is the point

(

h

,

k

)

.

Substitute the vertex into the equation:

y

=

a

(

x

−

0

)

2

+

0

Simplify:

y

=

a

x

2

The distance, f, from the vertex to the focus is,

f

=

8

; this allows use to find the value of

a

, using the equation:

a

=

1

4

f

a

=

1

32

The equation of the parabola is:

y

=

1

32

x

2

Step-by-step explanation:

7 0

3 years ago

Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cot θ= -6/7 . Find the exact values of

Llana [10]

Answer:

Part 1) $csc(\theta)=-\frac{\sqrt{85}}{7}$

Part 2) $sin(\theta)=-\frac{7}{\sqrt{85}}$ or $sin(\theta)=-\frac{7\sqrt{85}}{85}$

Part 3) $tan(\theta)=-\frac{7}{6}$

Part 4) $cos(\theta)=\frac{6}{\sqrt{85}}$ or $cos(\theta)=\frac{6\sqrt{85}}{85}$

Part 5) $sec(\theta)=\frac{\sqrt{85}}{6}$

Step-by-step explanation:

we know that

The angle theta lie on the IV Quadrant

so

sin(θ) is negative

cos(θ) is positive

tan(θ) is negative

sec(θ) is positive

csc(θ) is negative

step 1

Find the value of csc(θ)

we know that

$1+cot^{2}(\theta)=csc^{2}(\theta)$

we have

$cot(\theta)=-\frac{6}{7}$

substitute

$1+(-\frac{6}{7})^{2}=csc^{2}(\theta)$

$1+\frac{36}{49}=csc^{2}(\theta)$

$\frac{85}{49}=csc^{2}(\theta)$ rewrite

$csc(\theta)=-\frac{\sqrt{85}}{7}$ ----> remember that is negative

step 2

Find the value of sin(θ)

we know that

$csc(\theta)=\frac{1}{sin(\theta)}$

we have

$csc(\theta)=-\frac{\sqrt{85}}{7}$

therefore

$sin(\theta)=-\frac{7}{\sqrt{85}}$

or

$sin(\theta)=-\frac{7\sqrt{85}}{85}$

step 3

Find the value of tan(θ)

we know that

$tan(\theta)=\frac{1}{cot(\theta)}$

we have

$cot(\theta)=-\frac{6}{7}$

therefore

$tan(\theta)=-\frac{7}{6}$

step 4

Find the value of cos(θ)

we know that

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$sin(\theta)=-\frac{7}{\sqrt{85}}$

substitute

$(-\frac{7}{\sqrt{85}})^{2}+cos^{2}(\theta)=1$

$\frac{49}{85}+cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{49}{85}$

$cos^{2}(\theta)=\frac{36}{85}$

$cos(\theta)=\frac{6}{\sqrt{85}}$ ------> the cosine is positive

or

$cos(\theta)=\frac{6\sqrt{85}}{85}$

step 5

Find the value of sec(θ)

we know that

$sec(\theta)=\frac{1}{cos(\theta)}$

we have

$cos(\theta)=\frac{6}{\sqrt{85}}$

therefore

$sec(\theta)=\frac{\sqrt{85}}{6}$ ----> is positive

8 0

4 years ago

Pls help ion know how to this jaunt

Sedaia [141]

Answer: a very straight forward b 2 4 and 6 c 9

Step-by-step explanation:

7 0

2 years ago