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ruslelena [56]
3 years ago
8

Andrea's dad is 57 kilograms heavier than Andrea. Andrea weighs 34 kilograms. how much does Andrea's dad weigh?

Mathematics
2 answers:
lys-0071 [83]3 years ago
6 0
Too solve this, you just do 57 + 34 which is 91.
exis [7]3 years ago
6 0
All you do is 57+34= 91 because Andreas dad is 57 km MORE THAN her.
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Consider the function f(x) = ex and the function g(x), which is shown below. How will the graph of g(x) differ from the graph of
SIZIF [17.4K]

Answer: Option C

Step-by-step explanation:

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If b the graph moves vertically down

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Sec squared 55 - tan squared 55
sergejj [24]

<u>Answer: </u>

sec squared 55 – tan squared 55  = 1

<u>Explanation:</u>

Given, sec square 55 – tan squared 55

We know that,

\sec \Theta=\frac{\text {hypotenuse}}{\text {base}}

And,

\tan \theta=\frac{\text { perpendicular }}{\text { base }}

where Ө is the angle

Substituting the values

\left(\frac{\text {hypotenuse}}{\text {base}}\right)^{2}-\left(\frac{\text { perpendicular }}{\text {base}}\right)^{2}

Solving,

\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}

According to Pythagoras theorem,

\text { (hypotenuse) }^{2}-\text { (perpendicular) }^{2}=(\text { base })^{2}

Putting this in the equation;

squared 55 - tan squared 55 =

\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}=\frac{(\text {base})^{2}}{(\text {base}) *(\text {base})}=1

Therefore, sec squared 55 – tan squared 55 = 1

6 0
3 years ago
Given T5 = 96 and T8 = 768 of a geometric progression. Find the first term,a and the common ratio,r.
Alona [7]

Answer:

Of the given geometric sequence, the first term a is 6 and its common ratio r is 2.

Step-by-step explanation:

Recall that the direct formula of a geometric sequence is given by:

\displaystyle T_ n = ar^{n-1}

Where <em>T</em>ₙ<em> </em>is the <em>n</em>th term, <em>a</em> is the initial term, and <em>r</em> is the common ratio.

We are given that the fifth term <em>T</em>₅ = 96 and the eighth term <em>T</em>₈ = 768. In other words:

\displaystyle T_5 = a r^{(5) - 1} \text{ and } T_8 = ar^{(8)-1}

Substitute and simplify:

\displaystyle 96 = ar^4 \text{ and } 768 = ar^7

We can rewrite the second equation as:

\displaystyle 768 = (ar^4) \cdot r^3

Substitute:

\displaystyle 768 = (96) r^3

Hence:

\displaystyle r = \sqrt[3]{\frac{768}{96}} = \sqrt[3]{8} = 2

So, the common ratio <em>r</em> is two.

Using the first equation, we can solve for the initial term:

\displaystyle \begin{aligned} 96 &= ar^4 \\ ar^4 &= 96 \\ a(2)^4 &= 96 \\ 16a &= 96 \\ a &= 6 \end{aligned}

In conclusion, of the given geometric sequence, the first term <em>a</em> is 6 and its common ratio <em>r</em> is 2.

7 0
3 years ago
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