Pls help its past due 11 days

What is the distance between 182 and 265 on a number line?

adoni [48]

265-182=83
the answer is b.

6 0

3 years ago

Read 2 more answers

Each drink cost d in dollars

zhenek [66]

There is no picture what do you mean

6 0

2 years ago

An isosceles triangle has an angle that measures 38degrees. What measures are possible for the other two angles? Choose all that

Sphinxa [80]

Answer:

71° and 71°.

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Step-by-step explanation:

The triangle having two sides equal along with one different side is called an isosceles triangle.

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So, Let us assume the other sides (two equal sides) of the isosceles triangle as x. As the different side is already given in the question.

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We know that,

The sum of all angles of a triangle is 180°

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So, According to the question :

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${\longrightarrow \it\qquad { \ { { 38 \: }^{ \circ} + x + x = 180{}^{ \circ} }}}$

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${\longrightarrow \it\qquad { \ { { 38 \: }^{ \circ} + 2x = 180{}^{ \circ} }}}$

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${\longrightarrow \it\qquad { \ { 2x = 180{}^{ \circ} - { 38 \: }^{ \circ}}}}$

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${\longrightarrow \it\qquad { \ { 2x = 142{}^{ \circ} }}}$

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${\longrightarrow \it\qquad { \ { x = \dfrac{142{}^{ \circ}}{2} }}}$

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${\longrightarrow \it\qquad { \pmb { x = 71^{ \circ} }}}$

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Therefore,

The measure of the two other angles of the isosceles triangle are 71° and 71°.

3 0

2 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$