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3 years ago

7

Two of the angles in a triangle measure 41° and 61° What is the measure of the third angle?

1 answer:

professor190 [17]3 years ago

5 0

Answer:

78 degrees

Step-by-step explanation:

All inner angles of a triangle make a sum of 180 degrees. Given 41 and 61 add up to 102, that leaves 78 degrees left to make 180

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The probability that a certain kind of component will survive a shock test is 3/4. Find the probability that exactly 2 of the ne

Marina86 [1]

Answer:

Therefore the required probability is $=\frac{27}{128}$

Step-by-step explanation:

The probability of success is $\frac{3}{4}$

The number of trial = 4

X= the items survive out of 4

$P(x=r)=^nC_rq^{n-r}p^r$ p =the probability of success and q = the probability failure.

p= $\frac{3}{4}$ and $q=(1-\frac{3}{4})=\frac{1}{4}$

$\therefore P(X=2)=^4C_2(\frac{1}{4} )^2(\frac{3}{4} )^2$

$=\frac{4!}{2!2!} (\frac{1}{16} )(\frac{9}{16} )$

$=\frac{27}{128}$

Therefore the required probability is $=\frac{27}{128}$

3 0

3 years ago

9x2 - 6xy<br> Factorise Fully

tamaranim1 [39]

Remember distributive property, but reverse it
ab-ac=a(b-c)
a is a common factor

find common factors
factor each
9x^2=3*3*x*x
6xy=2*3*x*y
common factor is 3*x or 3x

9x^2-6xy=3x(3x)-3x(2y)=3x(3x-2y)

factored form is 3x(3x-2y)

4 0

3 years ago

Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

6 0

3 years ago