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3 years ago

Your Turn: Using the figure, a

Mathematics

1 answer:

Hatshy [7]3 years ago

8 0

We are given:
$\frac{a}{c} = \frac{d}{a}$

First step is to get rid of fractions. We start by multiplying both sides by a.
$\frac{a}{c} = \frac{d}{a} /*a \\$

Now we cancel a on right side and we multiply both sides by c.
$\frac{ a^{2}}{c} = \frac{da}{a} /*c \\ \frac{ a^{2}c}{c} = dc$

After canceling c on the right side we are left with:
$a^{2} = dc or a^{2} = cd$

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Is this a function ? Yes or no and why or why not

never [62]

Answer:

Yes

Step-by-step explanation:

Yes it is a function, because graph is intersecting y - axis only once.

3 0

2 years ago

Please help! 10 ∑ 2i-9 i=3

pentagon [3]

$\sum\limits_{i=3}^{10}(2i-9)=2(3)-9+2(4)-9+2(5)-9+2(6)-9+2(7)-9\\\\+2(8)-9+2(9)-9+2(10)-9\\\\=6-9+8-9+10-9+12-9+14-9+16-9+18-9+20-9\\\\=90-72=18$

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3 years ago

If an arithmetic progression satisfies U4=7 , U8=15 .Find U10

Alja [10]

Answer:

U10 = 19

Step-by-step explanation:

The sequence is $Ux=(x*2)-1$

x is the coefficient to U.

So, $U10=(10*2)-1$

Which means $U10=20-1$

Which means $U10=19$

And there is your answer.

Hope this helps. Have a nice day.

6 0

2 years ago

What is the value of the expression (2)^6? pls help

Ronch [10]

Answer:

Step-by-step explanation:

2^6, or 2⁶, is two to the power of six.

Basically you multiply the bottom number (in this case 2) by itself as many times as the the number in the top right (in this case six)

2^6=2*2*2*2*2*2=64

8 0

2 years ago

Read 2 more answers

How many pairs of consecutive natural numbers have a product of less than 40000? I am in 5th grade. This is supposed to be easy

Ugo [173]

Answer:

There are 199 pairs of consecutive natural numbers whose product is less than 40000.

Step-by-step explanation:

We notice that such statement can be translated into this inequation:

$n \cdot (n+1) < 40000$

Now we solve this inequation to the highest value of $n$ that satisfy the inequation:

$n^{2}+n < 40000$

$n^{2}+n -40000$

The Quadratic Formula shows that roots are:

$n_{1,2} = \frac{-1\pm\sqrt{1^{2}-4\cdot (1)\cdot (-40000)}}{2\cdot (1)}$

$n_{1,2} = -\frac{1}{2}\pm \frac{1}{2} \cdot \sqrt{160001}$

$n_{1} = -\frac{1}{2}+\frac{1}{2}\cdot \sqrt{160001}$

$n_{1} \approx 199.501$

$n_{2} = -\frac{1}{2}-\frac{1}{2}\cdot \sqrt{160001}$

$n_{2} \approx -200.501$

Only the first root is valid source to determine the highest possible value of $n$ , which is $n_{max} = 199$ . Each natural number represents an element itself and each pair represents an element as a function of the lowest consecutive natural number. Hence, there are 199 pairs of consecutive natural numbers whose product is less than 40000.

6 0

2 years ago