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2 years ago

13

In the word FLASHLIGHT what is the ratio of vowels to total letters? m.

2 answers:

Leviafan [203]2 years ago

4 0

2:10
2 vowels
10 total letters

Ksivusya [100]2 years ago

3 0

Answer:

The ratio is 2:10 or 2 vowels to 10 total letters ;)

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Ahmad has $x. Abu has $5 more than Ahmad. Razak has twice as much as Abu. Together they have $175. What is the value of x?

Alisiya [41]

Abu X + 5

Razak 2x + 10

Ahmad X

Together 4x + 15

175 - 15 = 160

160 / 4 = 40

X = 40

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2 years ago

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(-4,-6) and (3,-7) find the distance between each pair of points

Vlad [161]

Answer:

$5\sqrt{2}$

Step-by-step explanation:

distance between two points:

$d = \sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2}}$

we have:

(-4, -6), (3, -7)

$x_{1} = -4$

$y_{1} =-6$

$x_{2} = 3$

$y_{2} =-7$

so we have:

$d =\sqrt{(3-(-4))^{2}+ (-7-(-6))^{2}}\\\\d =\sqrt{(7)^{2}+ (-1)^{2}}\\\\d = \sqrt{49+ 1}\\\\d=\sqrt{50}\\\\d=5\sqrt{2}$

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2 years ago

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Convert the polar equation to rectangular form r=2/1+sin( theta)

Nataly_w [17]

$r=\dfrac2{1+\sin\theta}$
$r+r\sin\theta=2$
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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

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

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

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3 years ago

Hannah does push-ups in sets of 3.She did 10 sets pushups today.Then her coach asked her to do 15 more push-ups.How many push-up

Lana71 [14]

10x3= 30
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3 years ago

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