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3 years ago

I need help on this please

Mathematics

2 answers:

sergiy2304 [10]3 years ago

8 0

Answer:

81 Degrees/Base Angle

Step-by-step explanation:

x+x+18=180 1. Add x+x= 2x. 2. Subtract 18 from each side, which leaves you with 2x=162. 3. Divide each side by 2, that gives you a final answer of x=81

allochka39001 [22]3 years ago

5 0

Answer:

I cant help but if you look up base calculators calc for vertext you justr plug in the numbers and it gives you the answer I can prolly get it for you in a sec

Step-by-step explanation:

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My calculator wasn’t working for this I’m not sure how to solve it but help would be coolio

ziro4ka [17]

We know that:

$_nP_k = \dfrac{n!}{(n-k)!}$

so in this case $n = 5$ , $k=3$ and:

$_5P_3=\dfrac{5!}{(5-3)!}=\dfrac{5!}{2!}=\dfrac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=\dfrac{120}{2}=\boxed{60}$

4 0

2 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

2 years ago

What is the cube root of -1,000p^12q^3

White raven [17]

I hope this helps :)

5 0

2 years ago

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Write the trinomial as a square of a binomial<br> x^2+2xy+y^2

vitfil [10]

Answer:

$\large\boxed{x^2+2xy+y^2=(x+y)^2}$

Step-by-step explanation:

$(x+y)^2=(x+y)(x+y)\\\\\text{Use FOIL}\ (a+b)(c+d)=ac+ad+bc+bd\\\\=(x)(x)+(x)(y)+(y)(x)+(y)(y)=x^2+xy+xy+y^2=x^2+2xy+y^2$

3 0

3 years ago

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Which equation(s) have x = –3 as the solution? log3(2x + 15) = 2 log5(8x + 9) = 2 log4(–20x + 4) = 3 logx81 = 4

baherus [9]

Answer:

log_3(2x+15)=2 and log_4(-20x+4)=3

Step-by-step explanation:

Plug it in and see!

$log_3(2x+15)=2\\log_3(2(-3)+15)=2\\log_3(-6+15)=2\\log_3(9)=2\\\text{ This is a true equation because } 3^2=9\\\\\\log_5(8(-3)+9)=2\\log_5(-24+9)=2\\log_5(-15)=2\\\\\text{ Not true because you cannot do log of a negative number }\\\\log_4(-20(-3)+4)=3\\log_4(64)=3\\\text{ this is true because } 4^3=64\\\\\\log_x (81)=4\\log_{-3}(81)=4\\\text{ the base cannot be negative }\\\\\\\\\text{ There is only two options here} \\\\log_3(2x+15)=2 \text{ and } log_4(-20x+4)=3$

6 0

3 years ago

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