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3 years ago

Help pls I’ll give extra points

Mathematics

2 answers:

lakkis [162]3 years ago

6 0

1. shrill
2. luscious
3. Dreadful
4. Shiny

USPshnik [31]3 years ago

4 0

11) shrill
12)luscious
13)dreadful
14)shiny

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Write 55% as a decimal and as a fraction.

Sedaia [141]

The answer for this question is 0.55%

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What point on the line y=9x+8 is closest to the origin

Marat540 [252]

to find the point, find the location

realize that the shortest line that crosses through the origin and y=9x+8 is perpendicular to y=9x+8

perpendicular lines have slopes that multiply to get -1

y=9x+8 has a slope of 9

9*m=-1, m=-1/9

the slope of the mystery line is -1/9

since it passes through the origin, the y intercept is 0

y=(-1/9)x is the equation of the line from the point to the origin

find the intersection of this line and the original line

set them equal to each other

(-1/9)x=9x+8

multily both sides by 9

-x=81x+72

minus 81 both sides

-82x=72

divide both sides by -82

x=-36/41

find y

y=(-1/9)x

y=(-1/9)(-36/41)

y=4/41

the point is $(\frac{-36}{41},\frac{4}{41})$

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3 years ago

A 36 inch string is divided into 5 pieces. Two of the

icang [17]

X + x + 3(2+x)= 36
2x + 6 + 3x = 36
5x + 6 = 36
5x = 36 - 6
5x = 30
x = 6
Longer piece = 2 + 6 = 8inches

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3 years ago

How do you simplify (2xy2)3

pantera1 [17]

$(2xy^{2})^{3} = (2xy^{2})(2xy^{2})(2xy^{2})$
Now, we can just multiply it and obtain the result
$(2xy^{2})(2xy^{2})(2xy^{2})=(4x^{2}y^{4})(2xy^{2})=8x^{3}y^{6}$

I used law presented on this example
$2^{3}*2=2^{3}*2^{1}=2^{3+1}=2^{4}$

4 0

3 years ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

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

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

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

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

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3 years ago