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Sergio039 [100]
3 years ago
13

a snack that reginald just bought has 150 calories in 3/4 of a serving. how many calories per serving is this?

Mathematics
1 answer:
77julia77 [94]3 years ago
5 0
If there are 150 calories in 3/4 servings divide 150 by 3 and you get 50 ..you add 50 calories for the 1/4 servings which gives you 200 calories per serving
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I need help asap pls and thank you ;)
olga55 [171]

Answer:

\text{Length of AB is }\frac{ah}{a+h}

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP     (∵ corresponding angles)

∠NDC=∠NPM     (∵ corresponding angles)

By AA similarity rule,  ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\

KA(\frac{h}{x}-1)=AP

ND(\frac{h}{x}-1)=PD

Adding above two, we get

(KA+ND)(\frac{h}{x}-1)=(AP+PD)

⇒ (KN-AD)=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x^2}{h-x}

⇒ x^2=ah-ax-xh+x^2

⇒ x(h+a)=ah

⇒ x=\frac{ah}{a+h}

3 0
3 years ago
Triangle ABC is transformed to create triangle MNO and triangle PQR as described below
goldenfox [79]

Answer:

The correct answer is the second option

5 0
3 years ago
PLEASE PLEASE HELP ME! thank you so so much
viva [34]

Answer:

6π

Step-by-step explanation:

(135/360)×π×4²

(135/360)×π×16

=6π

6 0
3 years ago
Write the quadratic equation in standard form:<br> 7x + 8 + 2x2<br> 2x + 1 + x2
JulsSmile [24]

Answer:

3x^2 + 9x  +9

Step-by-step explanation:

Given

(7x + 8 + 2x^2) + (2x + 1 + x^2)

Required

The result in standard form

We have:

(7x + 8 + 2x^2) + (2x + 1 + x^2)

Remove brackets

7x + 8 + 2x^2 + 2x + 1 + x^2

Collect like terms

7x+ 2x  + 8 + 1 + 2x^2+ x^2

9x  + 9+ 3x^2

The standard form of a quadratic equation is:

ax^2 + bx + c

So, we have:

9x  + 9+ 3x^2

3x^2 + 9x  +9

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2 years ago
Math is hard can u help me
Mariana [72]
What do you need? I know many mathematical skills
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3 years ago
Read 2 more answers
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