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3 years ago

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:

If the graph of the function $g(x)=f(x) +b$ represents the transformations made to the graph of $y= f(x)$ then, by definition:

If $b> 0$ the graph moves vertically upwards.

If $b$ the graph moves vertically down

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3 years ago

Sec squared 55 - tan squared 55

sergejj [24]

Answer:

sec squared 55 – tan squared 55 = 1

Explanation:

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

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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 Tₙ is the nth term, a is the initial term, and r is the common ratio.

We are given that the fifth term T₅ = 96 and the eighth term T₈ = 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:

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Hence:

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So, the common ratio r is two.

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

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

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3 years ago