All categories

2 years ago

Diane has a temperature of 99°F. What is her temperature in Celsius?

Mathematics

1 answer:

xenn [34]2 years ago

8 0

Answer:

$37.\bar{2}\textdegree C$

Step-by-step explanation:

The formula for converting °F to °C is:

$(\textdegree F -32) \frac{5}{9}=\textdegree C$

We are given the temperature of 99°F.

=====================================

$(99\textdegree F -32) \frac{5}{9}=\textdegree C\\\rule{150}{0.5}\\(67) \frac{5}{9}= \textdegree C\\\\\boxed{37.\bar{2} = \textdegree C}$

=====================================

Diane's temperature in Celsius is 37.22...°C.

The 2 is recurring.

=====================================

Hope this helps!

You might be interested in

Find f(3) if f(x)=x^3 + 2x^2-x-1

Wittaler [7]

Answer:

$\underline{ \boxed{f(3) = 41}}$

Step-by-step explanation:

$if \to \: f(x)=x^3 + 2x^2-x-1 \\ then \: f(3) = {3}^{3} + 2( {3)}^{2} - 3 - 1 \\ f(3) = 27 + 18 - 4 \\ f(3) = 41$

4 0

2 years ago

F(x)=x^2-2x-3 (so I need the done in like 20 min)

maks197457 [2]

Vertex point: (1, -4)

X-intercepts: (-1, 0) , (3, 0)

Y-intercept: (0, -3)

Axis of symmetry: x = 1

This function will create a parabola that opens upwards. I’m not entirely sure what I was supposed to answer. Hopefully this covered it. I’ll attach an image of the parabola in case I missed anything.

7 0

3 years ago

Is 0.04 greater than 0.4

valentinak56 [21]

No
0.04 is actually smaller than 0.4 because the four is in the hundredths place which is smaller than the four in the tenths place of 0.4

7 0

3 years ago

Read 2 more answers

Help me on this please

zalisa [80]

Answer:

1. (x, y) → (x + 3, y - 2)

Vertices of the image

a) (-2, - 3)

b) (-2, 3)

c) (2, 2)

2. (x, y) → (x - 3, y + 5)

Vertices of the image

a) (-3, 2)

b) (0, 2)

c) (0, 4)

d) (2, 4)

3. (x, y) → (x + 4, y)

Vertices of the image

a) (-1, -2)

b) (1, -2)

c) (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the image

a) (1, -1)

b) (1, -2)

c) (2, -2)

d) (2, -4)

e) (3, -1)

f) (3, -3)

g) (4, -3)

h) (1, -4)

5. (x, y) → (x, y - 4)

Vertices of the image

a) (0, -2)

b) (0, -3)

c) (2, -2)

d) (2, -4)

6. (x, y) → (x - 1, y + 4)

Vertices of the image

a) (-5, 3)

b) (-5, -1)

c) (-3, 0)

d) (-3, -1)

Explanation:

To identify each IMAGE you should perform the following steps:

List the vertex points of the preimage (the original figure) as ordered pairs.

Apply the transformation rule to every point of the preimage

List the image of each vertex after applying each transformation, also as ordered pairs.

1. (x, y) → (x + 3, y - 2)

The rule means that every point of the preimage is translated three units to the right and 2 units down.

Vertices of the preimage Vertices of the image

a) (-5,2) (-5 + 3, -1 - 2) = (-2, - 3)

b) (-5, 5) (-5 + 3, 5 - 2) = (-2, 3)

c) (-1, 4) (-1 + 3, 4 - 2) = (2, 2)

2. (x,y) → (x - 3, y + 5)

The rule means that every point of the preimage is translated three units to the left and five units down.

Vertices of the preimage Vertices of the image

a) (0, -3) (0 - 3, -3 + 5) = (-3, 2)

b) (3, -3) (3 - 3, -3 + 5) = (0, 2)

c) (3, -1) (3 - 3, -1 + 5) = (0, 4)

d) (5, -1) (5 - 3, -1 + 5) = (2, 4)

3. (x, y) → (x + 4, y)

The rule represents a translation 4 units to the right.

Vertices of the preimage Vertices of the image

a) (-5, -2) (-5 + 4, -2) = (-1, -2)

b) (-3, -5) (-3 + 4, -2) = (1, -2)

c) (-1, -2) (-1 + 4, -2) = (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the preimage Vertices of the image

a) (-5, -2) (-5 + 6, -2 + 1) = (1, -1)

b) (-5, -3) (-5 + 6, -3 + 1) = (1, -2)

c) (-4, -3) (-4 + 6, -3 + 1) = (2, -2)

d) (-4, -5) (-4 + 6, -5 + 1) = (2, -4)

e) (-3, -2) (-3 + 6, -2 + 1) = (3, -1)

f) (-3, -4) (-3 + 6, -4 + 1) = (3, -3)

g) (-2, -4) (-2 + 6, -4 + 1) = (4, -3)

h) (-2, -5) (-2 + 3, -5 + 1) = (1, -4)

5. (x, y) → (x, y - 4)

This is a translation four units down

Vertices of the preimage Vertices of the image

a) (0, 2) (0, 2 - 4) = (0, -2)

b) (0,1) (0, 1 - 4) = (0, -3)

c) (2, 2) (2, 2 - 4) = (2, -2)

d) (2,0) (2, 0 - 4) = (2, -4)

6. (x, y) → (x - 1, y + 4)

This is a translation one unit to the left and four units up.

Vertices of the pre-image Vertices of the image

a) (-4, -1) (-4 - 1, -1 + 4) = (-5, 3)

b) (-4 - 5) (-4 - 1, -5 + 4) = (-5, -1)

c) (-2, -4) (- 2 - 1, -4 + 4) = (-3, 0)

d) (-2, -5) (-2 - 1, -5 + 4) = (-3, -1)

8 0

2 years ago

-2 2/3 divided by 4/3

Vedmedyk [2.9K]

Answer:

-2

Step-by-step explanation:

$Rule(s): a\frac{b}{c} = \frac{a \cdot c + b}{c}, \frac{-a}{b}=-\frac{a}{b}, \frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot \:d}{b\cdot \:c} \\2\frac{2}{3}=\frac{2\cdot 3+2}{3}\\=\frac{8}{3}\\=\frac{-\frac{8}{3}}{\frac{4}{3}}\\=-\frac{\frac{8}{3}}{\frac{4}{3}}\\=\frac{-8\cdot \:3}{3\cdot \:4}\\= \frac{-24}{12}\\= -2$

8 0

1 year ago