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Alex777 [14]
3 years ago
11

·A nutritionist recommends that an adult male consume 55 grams of per day. It is acceptable for the fat intake to differ from th

is amount by at most 25 grams. Write and solve an absolute-value inequality to find the range of fat intake that is acceptable. Graph the solutions.​

Mathematics
1 answer:
baherus [9]3 years ago
4 0
This is the answer I hope
It helps

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RS = 12, ST = 2x, RT = 34
Harrizon [31]

Answer:

x=11

Step-by-step explanation:

34(RT) minus 12(RS)=22. 22 divided by 2 equals 11.

6 0
3 years ago
PLEASE HELP?!?!?!?!?!?!?!
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The surface area is 72 ft.^2.
8 0
2 years ago
P(x) = x + 1x² – 34x + 343<br> d(x)= x + 9
Feliz [49]

Answer:

x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > \frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1. Go ahead and simplify.

x=\frac{9}{d-1};\quad \:d\ne \:1.

Now, \mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}.

P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343.

Group the like terms... 1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343.

\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1} > 1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343.

Now for 1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}.

\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}.

Now for 33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}.

Thus we then get \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343.

Now we want to combine fractions. \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}.

\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2

\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}

\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}.

Expand 81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac.

-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1.

\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}

Therefore P=\frac{-297d+378}{\left(d-1\right)^2}+343.

Hope this helps!

5 0
3 years ago
The level of calcium in the blood of healthy young adults follows a normal distribution with mean $\mu=10$ milligrams per decili
Allisa [31]
<span>This indicates the blood calcium level is very low .This condition is known as hypocalcemia. The low calcium level may result from a problem with the parathyroid glands as well as from diet,kidney disorders,or certain drugs also causes low blood calcium level.</span>
5 0
2 years ago
WILL MARK BRAINLIEST.
charle [14.2K]

Answer:

option d

Step-by-step explanation:

The cross section of the rectangular prism is parallel to the base.

So, the length and width is congruent to the base. So, same length and width

3 0
3 years ago
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