Ask question

Ask question

All categories

3 years ago

8

Help please bdjdnhsj

1 answer:

Yuliya22 [10]3 years ago

7 0

Answer:

True

Step-by-step explanation:

You might be interested in

A circle with a 10 inch diameter is cut from a square piece of tin with 10 inch sides. To the nearest tenth of a square inch, fi

saveliy_v [14]

The area of the circle is

... A = (π/4)d² . . . . where d=10 in

The area of the square is

... A = d² . . . . . . . . where d=10 in

The amount of the square that is not in the circle is

... (1 -π/4)d² = 0.2146018d² ≈ 21.5 in² . . . for d=10 in

5 0

3 years ago

Determine if the segment lengths form a triangle. If so, would the triangle be acute, right, or obtuse? 4.1, 8.2, 12.2

goldfiish [28.3K]

Three line segments can form a triangle if the length of the longest segment is greater than the sum of the lengths of the shorter segments.

$4.1 + 8.2 \ \textgreater \ 12.2 \\
12.3 \ \textgreater \ 12.2 \\
true$
They can form a triangle.

Now if c is the length of the longest side and a and b are the lengths of the shorter sides, then:
- if $c^2=a^2+b^2$ , the triangle is right
- if $c^2\ \textless \ a^2+b^2$ , the triangle is acute
- if $c^2\ \textgreater \ a^2+b^2$ , the triangle is obtuse

$4.1^2+8.2^2=16.81+67.24=84.05 \\
12.2^2=148.84 \\ \\
148.84\ \textgreater \ 84.05$
The triangle is obtuse.

The answer is D.

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

4 years ago

M-m-1/2=1-m-2/3 ( will mark brainiest)

Naya [18.7K]

Answer:

Step-by-step explanation:

m= 1/2

this is answer ..........

hope it helps

pls mark my answer as the brainliest

5 0

4 years ago

Read 2 more answers

If a person is given $1 and invests that $1 for 100 years at 5%, what amount would they be able to withdraw from this account af

Oksana_A [137]

in 100 years they will be able to withdraw $131.50

1(1.05)^100

6 0

3 years ago