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3 years ago

What is the solution to this equation?

Mathematics

2 answers:

Morgarella [4.7K]3 years ago

7 0

Answer:

Step-by-step explanation:

-0.2(x-20)=44-x

Divide both sides by -0.2:

x-20=-220+5x

Subtract x from both sides:

-20=-220+4x

Add 220 to both sides:

200=4x

x=50, or answer choice C. Hope this helps!

Anna007 [38]3 years ago

7 0

x = 50

Isolate the variable by dividing each side by factors that don't contain the variable. x=50 this is helpful

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Answer:

for both its unlikely but I would need to know many green and red there were to be sure

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3 years ago

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xxTIMURxx [149]

Answer:

Therefore,

AD = 7 unit

AE = 600 unit

EF = 30 unit

FB = 5 unit

Ar(EFGH)= 210 unit²

Ar(ABCD)= 4445 unit²

Step-by-step explanation:

Given:

Consider a rectangle ABCD as shown in the figure below where,

Area (ABCD) = 635 × 7 = 4445

Therefore length and width of a rectangle will be

LENGTH =635 = AB

WIDTH = 7 = AD

To Find:

AE =?

EF = ?

FB = ?

Ar(EFGH)=?

Ar(ABCD) =?

Solution:

We know that area of the rectangle is given by

$\textrm{Area of rectangle}= LENGTH\times WIDTH$

$\textrm{Area of rectangle ABCD}= 635\times 7=4445\ unit^{2}$

Substituting the values we get

$\textrm{Area of rectangle AEHD}= AE\times 7\\\\4200=AE\times 7\\\therefore AE=\dfrac{4200}{7}=600$

Similarly for area rectangle FBCG,

$\textrm{Area of rectangle FBCG}= FB\times 7\\\\35=FB\times 7\\\therefore FB=\dfrac{35}{7}=5$

Now we have

AB = 635,

$AE + EF + FB = 635\\\\600+EF+5=635\\\\EF=635 -605=30\\\therefore EF = 30$

Similarly for area rectangle EFGH,

$\textrm{Area of rectangle EFGH}= EF\times 7\\\\=30\times 7=210\ isunit^{2}$

Therefore,

AD = 7 unit

AE = 600 unit

EF = 30 unit

FB = 5 unit

Ar(EFGH)= 210 unit²

Ar(ABCD)= 4445 unit²

5 0

3 years ago

Read 2 more answers

An elementary school is offering 3 language classes: one in Spanish, one in French,and one in German. The classes are open to an

Alona [7]

Answer:

A. 0.5

B. 0.32

C. 0.75

Step-by-step explanation:

There are

28 students in the Spanish class,
26 in the French class,
16 in the German class,
12 students that are in both Spanish and French,
4 that are in both Spanish and German,
6 that are in both French and German,
2 students taking all 3 classes.

So,

2 students taking all 3 classes,
6 - 2 = 4 students are in French and German, bu are not in Spanish,
4 - 2 = 2 students are in Spanish and German, but are not in French,
12 - 2 = 10 students are in Spanish and French but are not in German,
16 - 2 - 4 - 2 = 8 students are only in German,
26 - 2 - 4 - 10 = 10 students are only in French,
28 - 2 - 2 - 10 = 14 students are only in Spanish.

In total, there are

2 + 4 + 2 + 10 + 8 + 10 +14 = 50 students.

The classes are open to any of the 100 students in the school, so

100 - 50 = 50 students are not in any of the languages classes.

A. If a student is chosen randomly, the probability that he or she is not in any of the language classes is

$\dfrac{50}{100} =0.5$

B. If a student is chosen randomly, the probability that he or she is taking exactly one language class is

$\dfrac{8+10+14}{100}=0.32$

C. If 2 students are chosen randomly, the probability that both are not taking any language classes is

$0.5\cdot 0.5=0.25$

So, the probability that at least 1 is taking a language class is

$1-0.25=0.75$

3 0

4 years ago

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never [62]

Answer:

Step-by-step explanation:

Given

$\frac{1}{2}$ x - $\frac{5}{4}$ + 2x = $\frac{5}{6}$ + x

Multiply through by 12 ( the LCM of 2, 4, 6 ) to clear the fractions

6x - 15 + 24x = 10 + 12x

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3 years ago

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What is the surface area of this figure?

photoshop1234 [79]

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4 years ago