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Naddik [55]
3 years ago
7

2.7 is greater than equal to b + 5? Solve

Mathematics
1 answer:
Semenov [28]3 years ago
7 0
The answer:   - 2.3 ≥ b ; which does not correspond with any of the answer choices; but most closely corresponds with: "Answer choice: [B]: b > -2.3 ." 
_____________
Explanation:
_________________
Assuming we have:
_______________________
2.7 is greater than <u><em>or</em></u> equal to "(b + 5)"; 
_______________________________
We would write:
_________________
→ 2.7 ≥  b + 5  ;
_________________
→ Subtract "5" from EACH side:
_________________
→ 2.7 − 5  ≥  b + 5 − 5

→ - 2.3 ≥ b ; which does not correspond with any of the answer choices; but most closely corresponds with: "Answer choice: [B]: b > -2.3 ."
____________________
You might be interested in
1. D=5.9 ft.
Shtirlitz [24]

\qquad\qquad\huge\underline{{\sf Answer}}♨

Here's the solution ~

As we know, we can calculate the circumference of a circle in terms of its diameter as :

\qquad \sf  \dashrightarrow \:c =  \pi d

where, c = circumference and d = diameter

And also, circumference of circle is terms of radius (r) is :

\qquad \sf  \dashrightarrow \:c =  2\pi r

Now, let's move on to questions ~

<h3>First </h3>

\qquad \sf  \dashrightarrow \:3.14 \times 5.9

\qquad \sf  \dashrightarrow \: \approx18.53 \: ft

<h3 />

・ .━━━━━━━†━━━━━━━━━.・

<h3>Second</h3><h3 /><h3 /><h3 /><h3>\qquad \sf  \dashrightarrow \:3.14 \times 3.2</h3>

\qquad \sf  \dashrightarrow \: \approx10.048 \: ft

・ .━━━━━━━†━━━━━━━━━.・

<h3>Third</h3>

\qquad \sf  \dashrightarrow \:3.14 \times 6.1

\qquad \sf  \dashrightarrow \: \approx19.15 \: ft

・ .━━━━━━━†━━━━━━━━━.・

<h3>Fourth</h3>

\qquad \sf  \dashrightarrow \:2×3.14 \times  3.7

\qquad \sf  \dashrightarrow \: \approx2×11.62  \: m

\qquad \sf  \dashrightarrow \: \approx23.24  \: m

・ .━━━━━━━†━━━━━━━━━.・

<h3>Fifth </h3>

\qquad \sf  \dashrightarrow \:2×3.14 \times  6.2

\qquad \sf  \dashrightarrow \: \approx2× 19.47 \: units

\qquad \sf  \dashrightarrow \: \approx38.94  \: m

・ .━━━━━━━†━━━━━━━━━.・

<h3>Sixth</h3>

\qquad \sf  \dashrightarrow \:2×3.14 \times  5.1

\qquad \sf  \dashrightarrow \: \approx2×16.01 \: ft

\qquad \sf  \dashrightarrow \: \approx \: 32.02m

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

7 0
2 years ago
Which one is right..
frez [133]

Answer:

the answer would be C

Step-by-step explanation:

8 0
3 years ago
A/20 + 14/15 = 9/15 <br><br> Can i get some help?
KiRa [710]
Method 1:

\dfrac{a}{20}+\dfrac{14}{15}=\dfrac{9}{15}\ \ \ \ |multiply\ both\ sides\ by\ 60\\\\60\cdot\dfrac{a}{20}+60\cdot\dfrac{14}{15}=60\cdot\dfrac{9}{15}\\\\3a+4\cdot14=4\cdot9\\\\3a+56=36\ \ \ \ |subtract\ 56\ from\ both\ sides\\\\3a=-20\ \ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{a=-\frac{20}{3}\to a=-6\frac{2}{3}}

Method 2.

\dfrac{a}{20}+\dfrac{14}{15}=\dfrac{9}{15}\ \ \ \ |subtract\ \dfrac{14}{15}\ form\ both\ sides\\\\\dfrac{a}{20}=-\dfrac{5}{15}\\\\\dfrac{a}{20}=-\dfrac{1}{3}\ \ \ \ |multiply\ both\ sides\ by\ 20\\\\\boxed{a=-\dfrac{20}{3}\to a=-6\frac{2}{3}}
3 0
3 years ago
Solve x + 5 &gt; 8.<br> The solution is ____
professor190 [17]
What do you mean?
Expand your question
4 0
3 years ago
Read 2 more answers
A father’s age now is three times the age that his son was 4 years ago. In
Mice21 [21]
Let ‘s’ be the son’s age 12 years ago.
Let ‘f’ be the father’s current age.

4 years ago, the son was:

s-4

So, his father is currently:

3(s-4)

=

3s-12

Therefore:

f = 3s-12

In twelve years, the son will be:

s+12

And the father will be:

f+12

This can also be written as:

3s-12+12 as the fathers younger age would be f = 3s+12

=

3s

So, we know that s+12 is half the fathers current age, meaning the father is currently 2(s+12) which is equivalent to 2s+24. Also, we know that the father is currently 3 times the sons age 12 years ago, so 3s (proven by the calculations we made above). Therefore, 2s+24=3s which means 24=s. We can then substitute this, and we will receive 24+12 = 36

Son’s current age: 36

We then substitute the son’s age 12 years ago into 2s+24 to give us the father’s age.

2(24)+24 = 72

Father’s current age: 72










7 0
2 years ago
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