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3 years ago

14

Find the error and the correct answerthx

2 answers:

irga5000 [103]3 years ago

4 0

Answer:

Only 6x6. It equals 36. This is the only thing I found was incorrect. Practice Multiplication!

Step-by-step explanation:

6x5= 30

30+6 = 36

pychu [463]3 years ago

3 0

6*6 isn't 6 it is 36 hope his helps

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-29 19/40 as a decimal

Mariulka [41]

19 divided by 40 is .475 so it’s -29.475

7 0

3 years ago

PLZ HELP ME BY THE END OF THE DAY

nlexa [21]

$3x^{2} +1=28$

Step 1 : subtract 1 from both sides

$3x^{2} =27$

Step 2 : Divide both sides by 3

$\frac{\frac{3x^{2} }{3} }{\frac{27}{3} }$

$x^{2} = 9$

Step 3 : Find the square root of 9

$\sqrt{9} = 3$

x= (-3,3)

remember negative × negative = positive

6 0

4 years ago

4 in.<br> 8 in.<br> 7 in.<br><br><br> Pls help

snow_lady [41]

Answer:

I assume you're asking for the volume, which is 224

Step-by-step explanation:

4x7x8=224

4 0

3 years ago

A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f

Andrews [41]

Answer:

$A = \frac{P}{r}\left( e^{rt} -1 \right)$

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

$\frac{dA}{rA+P} = dt$

Integration on both sides gives

$\int \frac{dA}{rA+P} = \int dt$

where $c$ is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

$\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c$

Therefore,

$\frac{1}{r} \ln |rA+P| = t+c$

Multiply both sides by $r.$

$\ln |rA+P| = rt+c_1, \quad c_1 := rc$

By taking exponents, we obtain

$e^{\ln |rA+P|} = e^{rt+c_1} \implies |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}$

Isolate $A$ .

$rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}$

Since $A = 0$ when $t=0$ , we obtain an initial condition $A(0) = 0$ .

We can use it to find the numeric value of the constant $c$ .

Substituting $0$ for $A$ and $t$ in the equation gives

$0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P$

Therefore, the solution of the given differential equation is

$A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)$

4 0

3 years ago

If cos A = k, then the value of the expression (sin A)(cos A)(tan A) is equivalent to:

sukhopar [10]

Answer:

1-k^2

Step-by-step explanation:

tan A= sin A/cos A

(sin A)(cos A)(sin A/cos A)=(sin A)^2

(sin A) ^2 + (cos A) ^2= 1

(sin A) ^2=1-(cos A) ^2

(sin A)^2= 1 - k^2

4 0

2 years ago

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