#5

Robin had 7/8 pound of butter when she started baking a cake. She used 1/4 pound of butter while baking. How much butter was lef

8_murik_8 [283]

Answer:

$\dfrac{5}{8}$ butter was left after she was done baking.

Step-by-step explanation:

Robin had 7/8 pound of butter when she started baking a cake. She used 1/4 pound of butter while baking.

We need to find how much butter was left after she was done baking.

It can be calculated as follows,

Left =

$=\dfrac{7}{8}-\dfrac{1}{4}\\\\=\dfrac{7}{8}-\dfrac{2}{8}\\\\=\dfrac{7-2}{8}\\\\=\dfrac{5}{8}$

Hence, $\dfrac{5}{8}$ butter was left after she was done baking

3 0

3 years ago

Write the equation of the line that passes through the points (8, –1) and (2, –5) in standard form, given that the point-slope f

dolphi86 [110]

Answer:

A linear equation in the standard form is written as:

y = a*x +b

where a is the slope and b is the y-intercept.

If we know that the line passes through the points (a, b) and (c, d), then the slope can be written as:

a = (d - b)/(c - a)

So we know that our line passes through the points (8, -1) and (2, -5)

then the slope will be:

a = (-5 - (-1) )/(2 - 8) = (-4/-6) = 2/3

Then our line is something like:

y = (2/3)*x + b

to find the value of b, we can use the fact that we know that our line passes through the point (2, -5)

this means that when x = 2, we must have y = -5

replacing these values in the equation we get:

-5 = (2/3)*2 + b

-5 = 4/3 + b

-5 - 4/3 = b

-15/3 - 4/3 = b

-19/3 = b

then the equation is:

y = (2/3)*x - 19/3

(in the question you wrote the point-slope form, but you can see that it does not work for the second point, so there may be a mistake there, as the slope is missing)

The actual equation in the point-slope form is:

y + 1 = (2/3)*(x - 8)

4 0

3 years ago

PLEASE HELP! I'M FAILING ALGEBRA CLASS AND I REALLY DON'T UNDERSTAND THIS!

motikmotik

Answer:

Yes

Step-by-step explanation:

85, 85, 90, 90, 95, 95, 100, 100, 150

Median: 95

Q1: (85+90)/2 = 87.5

Q3: 100

IQR: Q3 - Q1 = 100 - 87.5 = 12.5

1.5 × IQR = 18.75

Outliers are outside the interval:

[Q1 - 1.5×IQR, Q3 + 1.5×IQR]

Q3 + 1.5IQR = 100 + 18.75

= 118.75

Since 150 > 118.75, it is an outlier

7 0

3 years ago

PLEASE HELP ME I'M GIVING 20PTS AND MARKING BRAINLIEST!!!!

pishuonlain [190]

Using a trigonometric identity, it is found that the values of the cosine and the tangent of the angle are given by:

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

<h3>What is the trigonometric identity using in this problem?</h3>

The identity that relates the sine squared and the cosine squared of the angle, as follows:

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

In this problem, we have that the sine is given by:

$\sin{\theta} = \frac{1}{3}$

Hence, applying the identity, the cosine is given as follows:

$\cos^2{\theta} = 1 - \sin^2{\theta}$

$\cos^2{\theta} = 1 - \left(\frac{1}{3}\right)^2$

$\cos^2{\theta} = 1 - \frac{1}{9}$

$\cos^2{\theta} = \frac{8}{9}$

$\cos{\theta} = \pm \sqrt{\frac{8}{9}}$

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

The tangent is given by the sine divided by the cosine, hence:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

$\tan{\theta} = \frac{\frac{1}{3}}{\pm \frac{2\sqrt{2}}{3}}$

$\tan{\theta} = \pm \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

More can be learned about trigonometric identities at brainly.com/question/24496175

#SPJ1

5 0

2 years ago

PLEASE HELP I ONLY HAVE AN HOUR

Mariulka [41]

Answer:

A) -1/2

Honestly, I used Math/way to solve this for you. If you need help with anything else, let me know. :)

6 0

3 years ago