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mixas84 [53]
2 years ago
10

How many grams is equivalent to 16 milligrams?

Mathematics
2 answers:
natali 33 [55]2 years ago
7 0

Answer:

0.016

Step-by-step explanation:

Blababa [14]2 years ago
3 0
Your answer is the 3rd one (O.016)
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The community Relief Charity Group is earning 2.5% simple interest on the $2200 it invested in a savings account. It also earns
Over [174]

Answer:

$5,500 was invested in the insured bond fund

Step-by-step explanation:

First, let us calculate the interest earned on savings account

Simple interest = principal × rate × time

where:

principal = $2,200

rate = 2.5% = 2.5/100 = 0.025

time = 1 year

∴ simple interest = 2200 × 0.025 × 1 = $55

Total annual interest earned = interest earned from savings account + interests earned on insured bond fund = $308

∴ 308 = 55 + interests earned on insured bond fund

∴ interests earned on insured bond fund = 308 - 55 = $253

Finally, we know the simple interest earned on insured bond fund, the amount invested (principal) is calculated as follows:

simple interest = principal × rate × time

where:

simple interest = $253

rate = 4.6% = 0.046

time = 1 year

∴  253 = principal × 0.046 × 1

253 = principal × 0.046

∴principal = 253 ÷ 0.046 = 5500

Therefore $5,500 was invested in the insured bond fund

7 0
3 years ago
Solve and check.
Thepotemich [5.8K]
C/3 = 6 1/7

Multiply 3 to both sides:

C = 6 1/7 * 3

Convert 6 1/7 into an improper fraction:

C = 43/7 * 3

Multiply 3 to the numerator:

C = 129/7

Convert back into a mixed number:

C = 18 3/7
5 0
3 years ago
Read 2 more answers
Solve the differential equation by variation of parameters.<br><br> y''+y=secx
suter [353]

Answer:y=

Acosx+Bsinx +cosx ln(cosx)+x sinx

Step-by-step explanation:

given  equation y''+y=secx

auxiliary  equation

p^2+1=0\\p=\pm i

so CF is y=Acosx+Bsinx

now

y_1(x)=cosx \ \ \ \ y_2(x)=sinx\\{y_1}'(x)=-sinx \ \ \ \ {y_2}'(x)=cosx

using wronskian formula

W=\begin{vmatrix}cosx &sinx \\ -sinx & cosx\end{vmatrix}

          =cos^2x+sin^2x=1

now f(x)=secx

u=-\int \frac{f(x)y_2(x)}{W(x)}dx \ and \ v=\int \frac{f(x)y_1(x)}{W(x)}dx

u=-\int \frac{secx\times sinx}{1}dx \ and \ v=\int \frac{secx\times cosx}{1}dx

u=-\int tanx dx \ and \ v=\int {1}dx

u=ln(cosx) \ \ \ and \ \ v=x

now particular integrals are

PI=cosx ln(cosx)+x sinx

total solution

y= C.F+P.I

y=Acosx+Bsinx +cosx ln(cosx)+x sinx

8 0
3 years ago
Dividing 3 digits by 2 digits lesson 3.6
eimsori [14]
355÷5=



__71________
5(355
35 l
0 5
5
0
7 0
3 years ago
Multiply: (sqrt10 +2 sqrt8)(sqrt10-2 sqrt8)
vesna_86 [32]

Answer:

(√10 +2√8)(√10 -2√8)=

(10 -8√5 + 8√5 -32)

10+0-32

10-32

= -22

Hope this helps.

6 0
3 years ago
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