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denis23 [38]
3 years ago
12

A simplifie 6 × ( - 24 ) ÷ ( -4 )​

Mathematics
1 answer:
kumpel [21]3 years ago
8 0

\huge\textbf{Hey there!}

\huge\boxed{\mathsf{6\times(-24)\div (-4)}}\\\huge\boxed{= \mathsf{\dfrac{6(-24)}{-4}}}\\\huge\boxed{\mathsf{= \dfrac{\bold{-144}}{-4}}}\\\huge\boxed{\mathsf{\dfrac{-144\div-4}{-4\div-4}}}\\\huge\boxed{= \mathsf{\dfrac{\bold{36}}{1}}}\\\huge\boxed{\mathsf{= 36\div1}}\\\huge\boxed{\mathsf{= 36}}\\\\\\\huge\text{\bf Therefore, your answer is: \boxed{\mathsf{36}}}\huge\checkmark

\huge\text{\bf Good luck on your assignment and enjoy}\\\huge\text{\bf your day!}

~\frak{Amphitrite1040:)}

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X + (-3) = -7<br> X=<br> Answer here<br> I
tatiyna

Answer:

x=-4

Step-by-step explanation:

x-3=-7

x=-7+3

x=-4

3 0
4 years ago
Read 2 more answers
3x+2.2is equal to x=1.7
Leviafan [203]

Answer:

7.3 answer

Step-by-step explanation:

x=1.7

now

3x+2.2

3×1.7+2.2

7.3

7 0
3 years ago
Find the equation of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit.
leonid [27]

<u>Answer-</u>

The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are

12x-5y+14=0 \\ 12x-5y-14=0

<u>Solution-</u>

Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)

The applying the distance formula,

\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1

\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1

\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1

\Rightarrow 12h-5k-1=\pm 13

\Rightarrow 12h-5k=\pm 14

\Rightarrow 12h-5k=14,\ 12h-5k=-14

\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0

\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0

Two equations are formed because one will be upper from the the given line and other will be below it.

4 0
3 years ago
betty closes the nozzle and fills it completely with a liquid. She then opens the nozzle. If the liquid drips at the rate of 14
natta225 [31]

Answer:

4.71 minutes

Step-by-step explanation:

Incomplete question [See comment for complete question]

Given

Shape: Cone

r = 3 -- radius

h = 7 --- height

Rate = 14in^3/min

Required

Time to pass out all liquid

First, calculate the volume of the cone.

This is calculated as:

V = \frac{1}{3} \pi r^2h

This gives:

V = \frac{1}{3} * 3.14 * 3^2 * 7

V = \frac{1}{3} * 197.82

V = 65.94in^3

To calculate the time, we make use of the following rate formula.

Rate = \frac{Volume}{Time}

Make Time the subject

Time= \frac{Volume}{Rate }

This gives:

Time= \frac{65.94in^3}{14in^3/min}

Time= \frac{65.94in^3}{14in^3}min

Cancel out the units

Time= \frac{65.94}{14} min

Time= 4.71 min\\

3 0
3 years ago
Graph the line with slope 1/2 passing through the point (-4,4).
Yakvenalex [24]

Answer:

Step-by-step explanation:

------------

It's a straight line, so you need 1 additional point.

---

Plot (-4,4)

Slope 1/2 --> y increases 2 for each 1 of x increase.

Add 1 to x, and 2 to y ---> (-3,6)

4 0
3 years ago
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