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3 years ago

5

A patient receives milligrams (m) of medication per dose. Her doctor orders her medication to be decreased by 25 milligrams. How

much medication will she receive per dose? Write an algebraic expression

1 answer:

kolbaska11 [484]3 years ago

3 0

A (milligrams) - 25 milligrams = C milligrams

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*****50 POINTSSSS*****

defon

Answer:

<h3>Given</h3>

m∠REG = 78°
mAR = 46°
ER ≅ GA

<h3>Solution</h3>

m∠GAR = 180° - m∠REG = 180° - 78° = 102° (supplementary angles sum to 180°)
m∠TAR = 1/2mAR = 1/2(46°) = 23° (tangent chord angle is half the size of intercepted arc)
m∠GAN = 180° - (m∠TAR + m∠GAR) = 180° - (23° + 102°) = 55° (straight angle is 180°)
mAG = 2m∠GAN = 2(55°) = 110°
mRE = mAG = 110° (as ER ≅ GA)
mGE = 360° - (mAG + mAR + mRE) = 360° - (110° + 46° + 110°) = 94° (full circle is 360°)

6 0

2 years ago

What expression should replace A,B,C

ser-zykov [4K]

Answer:

A = 5x + 3y

B = 8x + 5y

C = 15x + 11y

Step-by-step explanation:

B = 6x + 2y + 2x + 3y

= 6x + 2x + 2y + 3y

= 8x + 5y

C = 8x + 5y + 7x + 6y

= 8x + 7x + 5y + 6y

= 15x + 11y

A = 7x + 6y - 2x - 3y

= 7x - 2x + 6y - 3y

= 5x + 3y

3 0

3 years ago

Can someone please answer. There is one problem. There's a picture thank you!

Whitepunk [10]

False I think, please don't quote me on that

7 0

2 years ago

Read 2 more answers

In a restaurant there are 3 choices for appetizers, 6 choices for entrees, and 4 choices for desserts. How many different meal c

Vlad1618 [11]

The correct answer is B. 72 because you can multiply 3*6*4 to represent every combination possible.

5 0

3 years ago

Wanda placed $100 in a savings account that pays 2% simple interest per year. After 6 months Wanda made another deposit of $100.

STatiana [176]

Answer:

$\$203.01$

Step-by-step explanation:

we know that

The simple interest formula is equal to

$A=P(1+rt)$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have

<em>After 6 months</em>

$t=6/12=0.5\ years\\ P=\$100\\r=2\%=2/100=0.02$

substitute in the formula above

$A_1=100(1+0.02*0.5)$

$A_1=100(1.01)$

$A_1=\$101$

<em>At the end of one year</em>

$t=6/12=0.5\ years\\ P=A_1+\$100=\$101+\$100=\$201\\r=2\%=2/100=0.02$

substitute in the formula above

$A_2=201(1+0.02*0.5)$

$A_2=201(1.01)$

$A_2=\$203.01$

8 0

3 years ago