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alisha [4.7K]
3 years ago
7

i know this is a lot but can someone pls help me with these 3 questions. i would very much appreciate it. c:

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
3 0
1.) 12 Hours

2.) 60%

3.) 24.3 lbs
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Find the diagonal of a rectangular frame which measures 77 in. by 36 in.
quester [9]

Answer:

Diagonal of a rectangular frame = 85 inch

Step-by-step explanation:

Given:

Length of rectangle = 77 inch

Width of rectangle = 36 inch

Find:

Diagonal of a rectangular frame

Computation:

Diagonal of a rectangle = √l² + b²

Diagonal of a rectangular frame = √77² + 36²

Diagonal of a rectangular frame = √5,929 + 1,296

Diagonal of a rectangular frame = √7,225

Diagonal of a rectangular frame = 85 inch

7 0
3 years ago
A survey was conducted to study the relationship between the annual income of a family and the amount of money the family spends
lisov135 [29]

Answer:

The correct option is d i.e. scatter plot

Step-by-step explanation:

The correct option is d i.e. scatter plot

scatter plot will be the best option to display the variation of expenditure with respect to annual income.

on one axis annual income is used and on the other expenditure of the family. hence for a particular change in annual income, an impact on expenditure will easily be predicted.

6 0
4 years ago
(6x-9)+5=2 can you solve this
ehidna [41]
(6x -9) + 5= 2
= -54 + 5 
=  2
7 0
3 years ago
Read 2 more answers
A shipping box in the shape of a rectangular prism has a volume of 18x^3+5x^2-2x. what are three expressions that can represent
Scrat [10]
Volume of the prism is given by:
Volume=length×width×height
but
V=<span>18x^3+5x^2-2x
V=x(18x^2+5x-2)
V=x(9x-2)(2x+1)
Hence the possible expression for the dimensions of the shipping box is x by (9x-2) by (2x+1)</span>
4 0
3 years ago
The angle θ1\theta_1 θ 1 ​ theta, start subscript, 1, end subscript is located in Quadrant II\text{II} II start text, I, I, end
melomori [17]

Answer:

\dfrac{3\sqrt{13}}{11}

Step-by-step explanation:

Given that the angle \theta_1  is located in Quadrant II; and

\cos(\theta_1)=-\dfrac{2}{11}

In Quadrant II, x is negative and y is positive.

\cos(\theta)=\dfrac{Adjacent}{Hypotenuse},\sin(\theta)=\dfrac{Opposite}{Hypotenuse}\\$Adjacent=-2\\Hypotenuse=11\\

To find \sin(\theta_1), we first determine the opposite angle of \theta_1.

This will be done using the Pythagoras theorem.

Hypotenuse^2=Opposite^2+Adjacent^2\\11^2=Opposite^2+(-2)^2\\Opposite^2=121-4=117\\Opposite=\sqrt{117}=3\sqrt{13}

Therefore:

\sin(\theta_1)=\dfrac{Opposite}{Hypotenuse}=\dfrac{3\sqrt{13}}{11}

6 0
3 years ago
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