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3 years ago

Terry weighs 40kg janice weighs 2 3/4 lg less than terry. what is their combined weight?

Mathematics

1 answer:

Liula [17]3 years ago

7 0

I hope this helps you

Janice 2 3/4= 2.4+3/4=11/4

40.11/4-40=110-40=70kg janice

Terry 40 kg

40+70=110

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What is the sale price to the nearest cent? $239 television; 10% discount

miv72 [106K]

Answer:

The sale price will be: $215.10

Step-by-step explanation:

We know that:

Sale price = Regular price × (100% - Discount %)

= Regular price × (100% - 10%)

= 239 × 90%

= 239 × 0.9

= 215. 10 $

Therefore, the sale price will be: $215. 10

3 0

3 years ago

PLEASE HELP ME I'M GIVING 20PTS AND MARKING BRAINLIEST!!!!

pishuonlain [190]

Using a trigonometric identity, it is found that the values of the cosine and the tangent of the angle are given by:

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

<h3>What is the trigonometric identity using in this problem?</h3>

The identity that relates the sine squared and the cosine squared of the angle, as follows:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

In this problem, we have that the sine is given by:

$\sin{\theta} = \frac{1}{3}$

Hence, applying the identity, the cosine is given as follows:

$\cos^2{\theta} = 1 - \sin^2{\theta}$

$\cos^2{\theta} = 1 - \left(\frac{1}{3}\right)^2$

$\cos^2{\theta} = 1 - \frac{1}{9}$

$\cos^2{\theta} = \frac{8}{9}$

$\cos{\theta} = \pm \sqrt{\frac{8}{9}}$

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

The tangent is given by the sine divided by the cosine, hence:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

$\tan{\theta} = \frac{\frac{1}{3}}{\pm \frac{2\sqrt{2}}{3}}$

$\tan{\theta} = \pm \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

More can be learned about trigonometric identities at brainly.com/question/24496175

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2 years ago

Eden just bought a trough in the shape of a rectangular prism for her horses. She needs to know what volume of water to add to t

notsponge [240]

The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Let w represent the width, hence:

length = w + 33, height = w - 13

Volume (V) = w(w + 33)(w - 13) = w³ + 20w² - 429w

V(w) = w³ + 20w² - 429w

Rate of change = dV/dw = 3w² + 40w - 429

When w = 38, dV/dw = 3(38)² + 40(38) - 429 = 5423

When w = 53, dV/dw = 3(53)² + 40(53) - 429 = 10118

Rate = 10118 - 5423 = 4695 in³/in

The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in

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4 0

2 years ago

The length of the rectangle is 5/6 of inches and the width of the rectangle is 2/15 of inches. What is the area of the rectangle

Inga [223]

Answer:

W = X in.

L = (3x - 6) in.

A = L*W = (3x-6)X = 45,

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3x(x-2) = 45,

x(x-2) = 15,

x^2 - 2x - 15 = 0

(x-5)(x+3) = 0,

x-5 = 0,

x = 5 in.

x+3 = 0,

x = -3.

Solution Set: x = 5, and x = -3.

Select + value of X:

x = 5

W = x = 5 in.

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Step-by-step explanation:

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2 years ago

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
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$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

3 years ago