All categories

2 years ago

How can you use four rods that measure 2 cm 5 cm 7 cm 9 cm to measure a length of one

1 answer:

5 0

<span>If you put the 9cm and 2cm rods together end-on-end, the new length is 11cm. Now if you repeat this with the 7cm and 5cm rods that new length is 12cm, so the difference between the two is now 1 cm. It does not seem very practical to walk around measuring items using the rods like this, so you could then make a small rod or use a strip of paper equal to the difference (1cm) as a standard measuring tool.</span>

You might be interested in

Please help and explain

Brilliant_brown [7]

Between the square roots of 7 and 8, because e^2 is equal to around 7.4, which is between 7 & 8. I hope this helps!

3 0

2 years ago

Please help ASAP thank you!!!!

mart [117]

Answer:

41. f⁻¹(x) = -9x + 4

43. m⁻¹(x) = ∛(x-2)/4

Step-by-step explanation:

41. y = (4-x)/9

swap x and y: x = (4-y)/9

solve y: 9x = 4-y

y = -9x + 4

45. y = 4x³+2

x = 4y³+2

4y³ = x-2

y³ = (x-2)/4

y = ∛(x-2)/4

Let you do 42 and 46 by yourself

5 0

2 years ago

The 13th term of a geometric sequence is 16, 384 and the first term is 4. What is the common ratio?

AleksAgata [21]

Answer:

$\large \boxed{2}$

Step-by-step explanation:

The formula for the nth term of a geometric sequence is

aₙ = a₁rⁿ⁻¹

In your geometric sequence, a₁ = 4 and a₁₃ = 16 384.

$\begin{array}{rcl}16384 & = & 4r^{(13 - 1)}}\\16384 & = & 4r^{12}\\4096 & = & r^{12}\\3.6124 & = & 12 \log r\\0.30102 & = & \log r\\r & = & 10^{0.30102}\\ & = & \mathbf{2}\\\end{array}\\\text{The common ratio is $\large \boxed{\mathbf{2}}$}$

Check:

$\begin{array}{rcl}16384 & = & 4(2)^{12}\\16384 & = & 4(4096)\\16384 & = & 16384\\\end{array}$

It checks.

7 0

2 years ago

Will give brainiest algebra 2

mars1129 [50]

The sum of squares of numbers is: 13

Step-by-step explanation:

Let x and y be two numbers

Then,

Difference of the squares of the numbers will be:

$x^2-y^2$

Product will be:

$xy$

Given identity is:

$(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2$

Given values are:

Difference of the squares of the numbers= $x^2-y^2=5$

Product of numbers = xy = 6

Putting the values in the identity

$(x^2+y^2)^2=(5)^2+[2(6)]^2\\=25+(12)^2\\=25+144\\=169$

As we have to only find x^2+y^2

Taking square root on both sides

$\sqrt{(x^2+y^2)^2}=\sqrt{169}\\x^2+y^2=13$

The sum of squares of numbers is: 13

Keywords: Identities

Learn more about identities at:

#LearnwithBrainly

7 0

3 years ago

Rewrite each expression as a polynomial in standard form.a)(x+i)(x-i)b)(x+5i)(x-5i)

Nitella [24]

Answer:

a) x² +1

b) x² +25

Step-by-step explanation:

a) (x+i)(x− i)

= x² - ( i ) x + ( i ) x - ( i)²

= x² - i² ∵ i² = -1

= x² - (-1)

= x² +1

b) (x+5 i)(x− 5i)

= x² - ( 5 i ) x + ( 5 i ) x - ( 5 i)²

= x² - 25 i² ∵ i² = -1

= x² - 25(-1)

= x² +25

we can also solve

using identity

(a + b)(a - b) = a² - b²

= (x+5 i)(x− 5i)

= x² - (5 i)²

= x² - 25 i²

= x² +25

8 0

3 years ago