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3 years ago

You have three quarters, two dimes, and five pennies in your pocket. You choose two coins without

Mathematics

1 answer:

Sveta_85 [38]3 years ago

4 0

It’s B because the quarter is big and you finna go for the big coin and then it’s the dime bc you not finna care

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Please help!!! I’ll give medals I’m stuck ;-(

AURORKA [14]

By inspection, LMNO is similar to PQRS.
For similar figures, corresponding sides are proportional, therefore

LM/PQ=MN/QR
Substituting values
4/2=6/c
Cross-multiply
4c=6*2=12
c=12/4=3

6 0

3 years ago

G(x)= x^2+1 and g(2k)=101, then k= ?

vlabodo [156]

I hope this helps you

4 0

4 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

the length of a rectangle is equal to triple the width . if the perimeter is 8 centimeters .what is the length and the width of

madam [21]

Answer:

Length = 3 cm

Width = 1 cm

Step-by-step explanation:

Let the length of rectangle be l and width of rectangle be w.

According to problem,

l = 3w {Length of rectangle is equal to triple the width}

And Perimeter,P = 8 cm

Since, P = 2 ( l + w )

or 8 = 2( l + w)

Plug l =3w in the above perimeter equation.

We get:

8 = 2( 3w + w)

8 = 2(4w)

8 = 8w

or w = 1 cm

Then length ,l = 3w =3 * 1 = 3 cm

Hence length of rectangle is 3cm and width of rectangle is 1cm.

3 0

3 years ago

Jaxons start up business makes a profit of 450 during the first month however the company records a profit of -60 per month for

Nataly_w [17]

Answer: $335

Step-by-step explanation: 4*-60=-240. $-240 is the profit for the next 4 months. Add this to 450. 450+(-240)= 450-240=210. Next add the final month's profit. 210+125= 335.

8 0

3 years ago