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2 years ago

How do I solve for d in the following equation? c(ad)+x=bd+ad(yz)

1 answer:

Klio2033 [76]2 years ago

8 0

Easy
remember associative property
a(bc)=(ab)c
so theerfor we can move dem around
also rembmer distributive property and how it can be expanded
ab+ac=a(b+c) therefor
ab+ac+ad=a(b+c+d)

(ca)d+x=bd+d(ayz)
minus (ca)d from both sides
x=db+d(ayz)-d(ca)
undistribute d
x=d(b+ayz-ca)
divide both sides by (b+ayz-ca)
$\frac{x}{b+ayz-ca}=d$
$d=\frac{x}{b+ayz-ca}$

You might be interested in

On which interval is the average rate of change for bicycle production the greatest? A) 1950 to 1960 B) 1960 to 1970 C) 1970 to

Alborosie

Answer:

D) 1980 to 2000

Step-by-step explanation:

Finding the average rate of change in each interval to determine the greatest one.

Production per interval

1950-1960 = $(21-11) \ million$ = 10 million

1960-1970 = $(41-21)\ million$ = 20 million

1970-1980 = $(71-41)\ million$ = 30 million

1980-2000 = $(161-71)\ million= 90$ million

Rate of change (1950-1960)= $\frac{10}{11} \times 100= 90.90\%$

Rate of change (1960-1970) = $\frac{20}{21} \times 100= 95.23\%$

Rate of change (1970-1980)= $\frac{30}{41} \times 100= 73.17\%$

Rate of change (1980-2000)= $\frac{90}{71} \times 100= 126.76\%$

∴ rate of change between 1980 to 2000 is 126.78%

7 0

2 years ago

PLEASE PLEASE PLEASE FIND THE UNKNOWNS WILL GIVE BRAIN TYSM!!

Wittaler [7]

Answer:

Sum of the opposite pair of a cyclic quadrilateral is 180°.

c= 180-102= 78°

c= 180-102= 78°d= 180-92= 88°

intercepted arc

=>2(102°)= 140+a°

a=64°

Simillarly

=> 2(d°)= a°+b°

b°= 112°

so ....

a=64°

a=64°b=112°

a=64°b=112°c=78°

a=64°b=112°c=78°d=88

hope it helps you ....

4 0

2 years ago

Businesses deposit large sums of money into bank accountsImagine an account with $10 million dollars in it.

adoni [48]

again, let's assume daily compounding means 365 days per year.

$~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ t=years\dotfill &1 \end{cases} \\\\\\ I = (10000000)(0.0212)(1)\implies \boxed{I=212000}$

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{daily, thus 365} \end{array}\dotfill &365\\ t=years\dotfill &1 \end{cases}$

$A=10000000\left(1+\frac{0.0212}{365}\right)^{365\cdot 1}\implies A\approx 10214256.88 \\\\\\ \underset{\textit{earned interest amount}}{10214256.88~~ - ~~10000000 ~~ \approx ~~ \boxed{214256.88}}$