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2 years ago

14

Wesley and Lincoln went fishing. Wesley caught a fish 15 3/16 inches long and Lincoln caught a fish 12 3/4 inches long. How many

inches longer was the fish that Wesley caught?

1 answer:

Arlecino [84]2 years ago

8 0

Answer:

2 7/16 inches longer

Step-by-step explanation:

15 3/16 - 12 3/4= 243/16 - 51/4 = 243/16 - 204/16= 39/16 inches longer

39/16=2 7/16 inches longer

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Tju [1.3M]

Answer:cam

Step-by-step explanation:

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3 years ago

Set up a right triangle model for this problem and solve by using a calculator. Follow the models above,

irakobra [83]

<em><u>Question:</u></em>

Set up a right triangle model for this problem and solve by using a calculator. Follow the models above,

A photographer stands 60 yards from the base of a lighthouse and observes that the angle between the ground and the top of the lighthouse is 41° How tall is the lighthouse?

<em><u>Answer:</u></em>

The height of lighthouse is 52.2 yards

<em><u>Solution:</u></em>

Given that photographer stands 60 yards from the base of a lighthouse and observes that the angle between the ground and the top of the lighthouse is 41 degree

The diagram is attached below

Consider a right angled triangle ABC

AB is the height of the lighthouse

BC is the distance between the base of a lighthouse and Photographer

As per given, BC = 60 yards

Angle between the ground and the top of the lighthouse is 41 degree

Angle ACB = 41 degree

To find: height of lighthouse i.e AB = ?

We know that,

$tan(\angle ACB) = \frac{Perpendicular}{Base}$

Here Base is BC and perpendicular is AB

$\tan 41^{\circ}=\frac{A B}{B C}$

Substituting the values,

$\begin{aligned}&\tan 41^{\circ}=\frac{A B}{60}\\\\&0.8692=\frac{A B}{60}\\\\&A B=0.8692 \times 60=52.157 \approx 52.2\end{aligned}$

Thus the height of lighthouse is 52.2 yards

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3 years ago

Using 4 equal-width intervals, show that the trapezoidal rule is the average of the upper and lower sum estimates for the integr

prisoha [69]

Split up the interval [0, 2] into 4 subintervals, so that

$[0,2]=\left[0,\dfrac12\right]\cup\left[\dfrac12,1\right]\cup\left[1,\dfrac32\right]\cup\left[\dfrac32,2\right]$

Each subinterval has width $\dfrac{2-0}4=\dfrac12$ . The area of the trapezoid constructed on each subinterval is $\dfrac{f(x_i)+f(x_{i+1})}4$ , i.e. the average of the values of $x^2$ at both endpoints of the subinterval times 1/2 over each subinterval $[x_i,x_{i+1}]$ .

So,

$\displaystyle\int_0^2x^2\,\mathrm dx\approx\dfrac{0^2+\left(\frac12\right)^2}4+\dfrac{\left(\frac12\right)^2+1^2}4+\dfrac{1^2+\left(\frac32\right)^2}4+\dfrac{\left(\frac32\right)^2+2^2}4$

$=\displaystyle\sum_{i=1}^4\frac{\left(\frac{i-1}2\right)^2+\left(\frac i2\right)^2}4=\frac{11}4$

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3 years ago

Which are the roots of the quadratic function f(q) = q2 – 125? Check all that apply.

s344n2d4d5 [400]

F(q)=0
so, q^2 - 125 = 0
(q)^2 - (5)^2 = 0
(q+5) (q-5) = 0
q = 5, -5

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3 years ago

Read 2 more answers

What is the degree of the monomial 15x^3y^9? (1 point)?

Contact [7]

Pretty sure it's 12.

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3 years ago