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2 years ago

Can anyone please help me with this RSM question? I have literally posted this like four times!

Mathematics

2 answers:

Ostrovityanka [42]2 years ago

5 0

Answer:

Angles in a triangle add up to 180.

ABG - x + 90 + ANGLE AGB = 180

Angles on a straight add up to 180.

CDE - 8x + ANGLE D + 3x = 180

Alterate angles - ATTACHMENT

ANGLE D = x

8x + x + 3x = 180

12x = 180

x = 180/12

zlopas [31]2 years ago

5 0

Answer:

x = 15

Step-by-step explanation:

Line FD is perpendicular to line EB. Line AC is perpendicular to line EB. Lines that are perpendicular to the same line are parallel. That makes lines FD and AC parallel. Angles EDF and C are corresponding angles of parallel lines cut by a transversal, so they are congruent. That makes m<C = m<EDF = 3x.

The sum of the measures of the angles of a triangle is 180 degrees.

Look at triangle ACD.

m<A + m<C + m<ADC = 180

x + 3x + 8x = 180

12x = 180

x = 15

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Practice multiplying numbers by powers of 10.

Juliette [100K]

Answer:

Multiply (400 + 20 + 3) x 10000

Add 4 zeros to the end of 423

Step-by-step explanation:

423 x 10,000 = Adding the amount of zero’s in 10000 to 423

8 0

3 years ago

Eight times the sum of 9 and some number is 192.

vovikov84 [41]

Answer:

Step-by-step explanation:

To solve this, divide 192 by 8:

192/9 = 24

Subtract 9 from 24:

24 - 9 = 15

Therefore, 8 times the sum of 9 and 15 is 192

3 0

3 years ago

Read 2 more answers

Molly tried to evaluate 93\times5193×5193, times, 51 using partial products. Her work is shown below.

Mrac [35]

Since molly's solution tally's with the given solution, hence <em>Molly's solution is correct.</em>

Given the working on a partial product of 93 and 51 carried out by Molly as shown:

$\begin{array}{llrr} &&93 \\ &&\underline{{}\times51} \\ &\blueD{\text{Step 1}}&\blueD{3}& \blueD{1\times3\text{ ones}}\\ &\greenD{\text{Step 2}}&\greenD{90}& \greenD{1\times 9\text{ tens}}\\ &\maroonD{\text{Step 3}}&\maroonD{150}& \maroonD{50\times 3\text{ ones}}\\ &\goldE{\text{Step 4}}&\underline{{}+\goldE{ 4{,}500}}& \goldE{50\times 9\text{ tens}}\\ &\purpleD{\text{Step 5}}&\purpleD{4{,}743}& \end{array}$

This partial product can also be solved as shown below:

$93 \times 51 = (90+3)\times (50+1)$

Applying the distributive law:

$93 \times 51 = 90(50) + 90(1) + 3(50) + 3(1)\\93 \times 51 =4500 + 90 + 150 + 3\\93 \times 51 =4500+240+3\\93 \times 51 =4740+3\\93 \times 51 =4743$

Since molly's solution tally's with the given solution, hence <em>Molly solution is correct.</em>

Learn more about partial product at: brainly.com/question/24716925

5 0

2 years ago

F<br> 2. What is the leading coefficient of the polynomial? - 7x2+ 3x -10<br> re

zhenek [66]

Answer:

The leading coefficient is -7

Step-by-step explanation:

The leading coefficient is the number in front of the highest power variable. In this case, the highest variable is x^2 and in front of it is -7. Therefore, the leading coefficient is -7

Answer: The leading coefficient is -7

5 0

2 years ago

Read 2 more answers

Help me with this...

marin [14]

i. 171

ii. 162

iii. 297

Solution,

n(U)= 630

n(I)= 333

n(T)= 168

i. Let n(I intersection T ) be X

$333 - x + x + 468 - x = 630 \\ or \: 333 + 468 - x = 630 \\ or \: 801 - x = 630 \\ or \: - x = 630 - 801 \\ or \: - x = - 171 \\ x = 171$

<h3>ii.n(only I)= n(I) - n(I intersection T)</h3><h3> = 333 - 171</h3><h3> = 162</h3>

<h3>iii. n ( only T)= n( T) - n( I intersection T)</h3><h3> = 468 - 171</h3><h3> = 297</h3>

<h3>Venn- diagram is shown in the attached picture.</h3>

Hope this helps...

Good luck on your assignment...

4 0

3 years ago