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3 years ago

What is the value of F in the formula F=9/5C+32 when C=30?

Mathematics

2 answers:

kherson [118]3 years ago

8 0

I don't know i think is C

Anni [7]3 years ago

5 0

Answer:

111.6

Step-by-step explanation:

F = 9/5 ( c+32)

C= 30

Then F= 9/5 (30 + 32)

= 9/5 x 62

= 1.8 x 62

= 111.6 degrees Fahrenheit

But this answer is not in your options provided

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Find the solution of y = -x - 3 for x = -2.

Neko [114]

Answer:

(-2,-1)

Step-by-step explanation:

y = -x - 3

you replace the x with the -2

y = -(-2) - 3

y=2-3

y=-1

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3 years ago

Denis, Vera, and Yash are rock climbers. Yash is connected to Vera by a

grin007 [14]

Answer:

$33.14\°$

Step-by-step explanation:

Let

Y ----> field of vision that Yash's camera would need

we know that

Applying the law of sines

$\frac{sin(Y)}{25}=\frac{sin(41\°)}{30}$

Solve for sin(Y)

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3 years ago

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3 years ago

An amusement park reduced it's admission price to $15.50 per day, but now charges $1.50 per ride. Mark has $26 to spend on admis

gtnhenbr [62]

15.50+1.50x =26
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3 years ago

Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge

Ede4ka [16]

Answer:

Part a) The quadratic function is $4x^{2} +20x-39=0$

Part b) The value of x is $1.5\ in$

Part c) The photo and frame together are $7\ in$ wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

$(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0$

Part b) What is the value of x?

Solve the quadratic equation $4x^{2} +20x-39=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem

we have

$4x^{2} +20x-39=0$

$a=4\\b=20\\c=-39$

substitute in the formula

$x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}$

$x=\frac{-20(+/-)\sqrt{1,024}} {8}$

$x=\frac{-20(+/-)32} {8}$

$x=\frac{-20(+)32} {8}=1.5\ in$ -----> the solution

$x=\frac{-20(-)32} {8}=-6.5\ in$

Part c) How wide are the photo and frame together?

$(4+2x)=4+2(1.5)=7\ in$

5 0

3 years ago