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hram777 [196]
2 years ago
14

Help help help hemp grko

Mathematics
1 answer:
Yuliya22 [10]2 years ago
5 0

Answer:

12

Step-by-step explanation:

x-4=2(x+8)/5

let's do my favorite strategy: distributive property!

x-4=(2x+16)/5

Now we can simplify.

x-4=(2/5)x+16/5

Let's solve for x!

Subtract 2/5 from both sides

3/5)x=36/5

Now multiply both sides by 5/3, the reciprocal.

You would get 12 as x!

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Which equation is a linear function?
Gwar [14]

Answer: y=3*2+2

Step-by-step explanation:

5 0
3 years ago
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
3 years ago
8.5+4(1-2.5k)=24.5 round to the nearest hundredth
frosja888 [35]
Hi there! 

Your question: 

8.5 + 49(1-2.5k)=24.5

My answer: 

8.5 + 4(1-2.5k) = 24.5 >>>>>>> first, you have to distribute

8.5 + 4 - 10k = 24.5 >>>>>>>> now, we are going to combine like terms.

12.5 - 10k = 24.5 >>>>>> now, we are going to subtract 12.5 on both sides 

-10k = 12 >>>>>>>>> finally, divide -10 on both sides

k = -1.2


Hope this helps you!


7 0
3 years ago
Read 2 more answers
Latrell and his cousin, Tisha, met at a state park to go for an all-day hike. Latrell drove 122 miles to get to the park, and it
aliina [53]

Answer:

to find the adverage you have to add the times together then divide them by 2 so 2 + 1.5 = 3.5/2= 1.75

Step-by-step explanation:

4 0
3 years ago
Help don't forget to put the work thank you
marysya [2.9K]

Answer:

<em>(J) </em>

Step-by-step explanation:

If the scale factor of dilation is greater than 1, the image is an enlargement.

If the scale factor of dilation is between 0 and 1, the image is a reduction.

6 0
3 years ago
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