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sweet-ann [11.9K]

4 years ago

6

A chef knows it will take two hours and fifteen minutes to prepare and cook a lunch. He starts preparing the lunch at a quarter

at a quarter to eleven in the morning. What time will lunch be ready

2 answers:

beks73 [17]4 years ago

7 0

Well simply minus fifteen minutes for each qauter

crimeas [40]4 years ago

3 0

It will be ready by 1 o clock :)

Hope this helps

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erastovalidia [21]

2 x 10 to the power of 4

3 0

3 years ago

Read 2 more answers

Can anybody help me with 1 and 2

Anit [1.1K]

1) $x \geq 96$
2) $x\ \textless \ 38$

Here, the variable $x$ can assume any value, and usually an inequality presented like this has $x$ as a stand-in for all real numbers. If we really wanted to make that fact clear, we could write the two inequalities like this:

$x \geq 96, x\in\mathbb{R}\\ x\ \textless \ 38, x\in\mathbb{R}$

Formally, you'd read $x\in\mathbb{R}$ as " $x$ is an element of the set of real numbers," which is a fancy way of saying that $x$ can be any real number.

8 0

3 years ago

Radian question please answer asap!!!

anygoal [31]

Answer: how about 90

Step-by-step explanation:

Its a 360 if you multiply 90×4 you get 360 every side is 90 degrees

5 0

3 years ago

Can you help me with this? 7

andrezito [222]

Answer:

65º

Step-by-step explanation:

The angle of a straight line is 180º, so ∠ABD=180º and ∠ABC=(180-6x)º
The sum of the interior angles of a triangle is 180, so (x+40)º+(3x+10)º+(180-6x)º=180
We can solve from there, x+40+3x+10+180-6x=180
Combine like terms, -2x+230=180
Subtract 230, -2x=-50
Divide by -2, x=25

m∠CAB=(x+40)º=(25+40)º=65º
m∠ABC=(180-6x)º=(180-150)º=30º
m∠BCA=(3x+10)º=(75+10)º=85º

4 0

3 years ago

Rina8888 [55]

To solve this we are going to use the formula for speed: $S= \frac{d}{t}$
where
$S$ is the speed
$d$ is the distance
$t$ is the time

Let $S_{l}$ be the speed of the boat in the lake, $S_{a}$ the speed of the boat in the river, $t_{l}$ the time of the boat in the lake, and $t_{a}$ the time of the boat in the river.

We know for our problem that <span>the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
</span> $S_{a}=S_{l}-2$
We also know that in the lake the boat<span> sailed for 1 hour longer than it sailed in the river, so:
</span> $t_{l}=t_{a}+1$
<span>
Now, we can set up our equations.
Speed of the boat traveling in the river:
</span> $S_{a}= \frac{6}{t_{a} }$
But we know that $S_{a}=S_{l}-2$ , so:
$S_{l}-2= \frac{6}{t_{a} }$ equation (1)

Speed of the boat traveling in the lake:
$S_{l}= \frac{15}{t_{l} }$
But we know that $t_{l}=t_{a}+1$ , so:
$S_{l}= \frac{15}{t_{a}+1}$ equation (2)

Solving for $t_{a}$ in equation (1):
$S_{l}-2= \frac{6}{t_{a} }$
$t_{a}= \frac{6}{S_{l}-2}$ equation (3)

Solving for $t_{a}$ in equation (2):
$S_{l}= \frac{15}{t_{a}+1}$
$t_{a}+1= \frac{15}{S_{l}}$
$t_{a}=\frac{15}{S_{l}}-1$
$t_{a}= \frac{15-S_{l}}{S_{l}}$ equation (4)

Replacing equation (4) in equation (3):
$t_{a}= \frac{6}{S_{l}-2}$
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$

Solving for $S_{l}$ :
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$
$(15-S_{l})(S_{l}-2)=6S_{l}$
$15S_{l}-30-S_{l}^2+2S_{l}=6S_{l}$
$S_{l}^2-11S_{l}+30=0$
$(S_{l}-6)(S_{l}-5)=0$
$S_{l}=6$ or $S_{l}=5$

We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.

3 0

4 years ago