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2 years ago

NEED HELP If you change 1 and half to Half Then how would you change half Into the same and amount

Mathematics

1 answer:

Karo-lina-s [1.5K]2 years ago

4 0

Answer:

You can't substitute baking soda with conditioner so don't add more conditioner becaue you'll not get fake snow. You should go and buy more baking soda.

Good Luck

Step-by-step explanation:

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Can someone help me with this problem? Thank you!

vesna_86 [32]

Answer: the answer is 180 $in^{2}$

Step-by-step explanation:

Area of a rectangle multiply length times base. 15 x 12 = 180. Hope this helps!

6 0

1 year ago

At a sport clubhouse the coach wants to rope off a rectangular area that is adjacent to the building. He uses the length of the

Furkat [3]

Answer:

w < 8 meters ( w must be greater than 0 because length cannot be 0)

Step-by-step explanation:

One side with building is 23 meters, the other opposite side also will be 23 meters (with rope).

Let width (remaining 2 sides) be "w", he has AT MOST 39 meters of rope, so we can write:

Rope Needed = 23 + 2w < 39

Simplifying:

$23 + 2w < 39\\2w < 39 - 23\\2w < 16\\w < 8$

The range of possible values of w is $w$ meters (of course w has to be greater than 0)

5 0

2 years ago

Mike and Julie surveyed students to determine how many had a grade of A in math. They recorded their findings for each survey in

grin007 [14]

20/4 = 5

40/8 = 5

60/12 = 5

80/16 = 5

Answer is 5

6 0

2 years ago

Read 2 more answers

The masses to the nearest kilogram of nine men were:

serious [3.7K]

Answer:

82 kg

Step-by-step explanation:

To find the mean of a set of values, we must add them up and divide by the number of values.

Step 1, adding the values:

$75+68+78+82+85+90+88+92+76=\\734$

Step 2, dividing by the # of terms:

There are 9 terms in total.

$\frac{734}{9} =\\81.556$

To the nearest kilogram, we can round 81.556 into 82.

The mean mass of the 9 men was $\fbox{82}$ kg.

<em>I hope this helps! Let me know if you have any questions :)</em>

5 0

2 years ago

Read 2 more answers

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

$g(y) = \frac{y-5}{y^2-3y+15}$

According to the quotient rule:

$d(\frac{f(y)}{h(y)}) = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}$

Applying the quotient rule:

$g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}$

The values for which g'(y) are zero are the critical points:

$g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10$

The critical values are y = 0 and y = 10.

5 0

2 years ago