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Radda [10]
3 years ago
8

Question 3

Mathematics
2 answers:
lana [24]3 years ago
5 0

Answer:

85cm^2

Step-by-step explanation:

<em>here's</em><em> your</em><em> </em><em>solution</em>

<em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>the </em><em>height</em><em> of</em><em> </em><em>traingle</em><em> </em><em>=</em><em> </em><em>5</em><em>c</em><em>m</em>

<em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>base </em><em>is </em><em>=</em><em> </em><em>3</em><em>4</em><em>c</em><em>m</em>

<em> </em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>area </em><em>of </em><em>traingle</em><em> </em><em>=</em><em> </em><em>1</em><em>/</em><em>2</em><em>*</em><em>base*</em><em>height</em>

<em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>,</em><em>to </em><em>find</em><em> out</em><em> </em><em>area </em><em>=</em><em> </em><em>1</em><em>/</em><em>2</em><em>*</em><em>3</em><em>4</em><em>*</em><em>5</em>

<em> </em><em> </em><em>=</em><em>></em><em>area </em><em>=</em><em> </em><em>8</em><em>5</em><em>c</em><em>m</em><em>^</em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>hope</em><em> it</em><em> helps</em>

<em> </em>

<em> </em><em> </em><em> </em>

Bond [772]3 years ago
5 0

Answer:

sorry just need points

Step-by-step explanation:

Just need points g

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Give the solution set for the inequality<br> 7x &lt; 7(x - 2) in interval notation.
eduard

<u>ANSWER:</u>

The solution set for the inequality 7x < 7(x - 2) is null set \varnothing

<u>SOLUTION:</u>

Given, inequality expression is 7x < 7 × (x – 2)

We have to give the solution set for above inequality expression in the interval notation form.

Now, let us solve the inequality expression for x.

Then, 7x < 7 × (x – 2)

7x < 7 × x – 2 × 7

7x < 7x – 14

7x – (7x – 14) < 0

7x – 7x + 14 < 0

0 + 14 < 0

14 < 0  

Which is false, so there exists no solution for x which can satisfy the given equation.

So, the interval solution for given inequality will be null set

Hence, the solution set is \varnothing

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3 years ago
Why do we have to change an improper fraction to a mixed fraction​
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Answer:

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Step-by-step explanation:

6 0
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9. Alicia Martin's savings account has a principle of $1,200. It earns 6% interest compounded quartly
zavuch27 [327]

Answer:

9) \$1236.27\,10)\,\$6451.07\, 11)\,\$10,152.87 \,12)\,\$907.95 \,13)\,\$4957.69

Step-by-step explanation:

9) Since Alicia Martin's savings earns 6% quarterly for two quarters then:

A=P(1+\frac{r}{n})^{nt} ⇒ Amount (A), Principle (P), rate (r) in decimal form, number of compoundings (n) a year and t, in year or its fractions.

A=P(1+\frac{r}{n})^{nt}\Rightarrow A=1200(1+\frac{0.06}{4})^{4*\frac{1}{2}}\Rightarrow A=\$1236.27

10) Aubrey Daniel's case:

A=P(1+\frac{r}{n})^{nt}\Rightarrow A=5725(1+\frac{0.04}{4})^{4*3}\Rightarrow A\approx \$6451.07

11) As for Angelo, similarly to Alicia.

A=P(1+\frac{r}{n})^{nt}\Rightarrow A=9855(1+\frac{0.06}{4})^{4*\frac{1}{2}}\Rightarrow A\approx \$10,152.87

12) Simpson's. For semiannual n=2

A=P(1+\frac{r}{n})^{nt}\Rightarrow A=860(1+\frac{0.055}{2})^{2*1}\Rightarrow A\approx \$907.95

13) Jana Lacey amount:

A=P(1+\frac{r}{n})^{nt}\Rightarrow A=4860(1+\frac{0.04}{4})^{4*\frac{1}{2}}\Rightarrow A\approx \$4957.69

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Answer:

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Step-by-step explanation:

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