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DENIUS [597]
3 years ago
15

6wholes and 1/4 ÷ 5wholes

Mathematics
1 answer:
Irina-Kira [14]3 years ago
4 0
Answer: 1 1/4      (1 whole and 1/4)

6 1/4 = 25/4
5 = 5/1

(25/4)/(5/1) = (25/4)*(1/5)
(25/4)*(1/5)
25/20 
1 5/20 = 1 1/4
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Hong hikes at least 1 hour but not more than 4 hours. She hikes at an average rate of 2.7 mph. The function f(t)=2.7t represents
Nimfa-mama [501]
C because its the most best answer.
6 0
4 years ago
3+6x129-834+23 divided by 20
Lemur [1.5K]

Answer:

- 1.7

Step-by-step explanation:

3 + 6 \times 129 - 834 + 23 =  - 34

- 34 \div 20 =  - 1.7

4 0
3 years ago
A food company originally sells cereal in boxes with dimensions 25 cm by 14 cm by 10 cm. To make more profit, the company decrea
hjlf

Answer:

1.75 cm

Step-by-step explanation:

We are given that the dimensions of the box as 25 cm by 14 cm by 10 cm.

Now, to increase the profit, the dimensions of the box by 'x' cm.

Thus, the new dimensions are (25-x) cm by (14-x) cm by (10-x) cm.

Further, it is given that the volume of the new box is 2,208 cm^{3}

As, Volume of the box = Length × Breadth × Height

i.e. 2208 = (25-x) × (14-x) × (10-x)

i.e. 2208 = (25-x) \times (140-24x-x^{2})

i.e. 2208=x^{3}-x^{2}-740x+3500

i.e. x^{3}-x^{2}-740x+1292=0

On solving this cubic equations, we get that, the solutions are x = 27.6, x = 1.75 and x = 26.8.

Since, the dimensions 25 cm by 14 cm by 10 cm are reduced by x cm.

Thus, if x = 27.6 or x = 26.8, then (10-x) = -17.6 or -16.8, which cannot be possible. So, x = 1.75 cm.

Hence, the dimensions were decreased by 1.75 cm.

3 0
3 years ago
Bru help please.... like r n
Bingel [31]

Answer: basically a rising straight line that goes through the origin

Step-by-step explanation:

5 0
3 years ago
Sin(θ) cos(3θ) + cos(θ) sin(3θ) = 0
pychu [463]

\displaystyle\\
\text{We use formula:}\\
\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+cos(\alpha)\sin(\beta)\\\\
Resolve:\\\\
\sin(\Theta)cos(3\Theta)+cos(\Theta)sin(3\Theta)=0\\\\
\sin(\Theta+3\Theta)=0\\\\
\sin(4\Theta)=0\\\\
\text{We have two solutions:}\\\\
4\Theta=0~~\Longrightarrow~~\Theta=\frac{0}{4}=0~~\Longrightarrow~~ \boxed{\Theta_1=0+2k\pi}\\\\
4\Theta=\pi~~\Longrightarrow~~\Theta=\frac{\pi}{4}~~\Longrightarrow~~ \boxed{\Theta_2=\frac{\pi}{4}+2k\pi}\\\\



5 0
3 years ago
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