All categories

2 years ago

Need Help Now! The tape diagram represents an equation. Write an equation to represent the image.

Mathematics

2 answers:

rjkz [21]2 years ago

7 0

Hi! From what I know, your answer should be 5n = 1.75.

This is because all 5 of those n's ultimately equal the large box of 1.75 on the top. Think of it as fractions. if we replaced 1.75 with 1, those n's would each be 1/5.

Fudgin [204]2 years ago

5 0

Answer:

1.75 = 5n

Step-by-step explanation:

hope this helps

You might be interested in

4x^2-4x=+1 by factoring

Alina [70]

Answer:

Below.

Step-by-step explanation:

4x^2 - 4x - 1 = 0

This will not factor.

But 4x^2 - 4x + 1 = 0 will:

= (2x - 1)(2x - 1) = 0

x = 0.5 (*2)

7 0

2 years ago

(a) Use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integral from 0 to pi/2 of sin(x)^

Sedbober [7]

Hello,

a)
$I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\

= [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\

=0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\

= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\
= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\

$
$I(1+n-1)= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
I= \dfrac{n-1}{n} *\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
$

b)
$\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\
= \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\
= \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\




$

$\int\limits^{ \frac{\pi}{2} }_0 {sin^{5}(x)} \, dx \\
= \dfrac{4}{5}*\dfrac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx = \dfrac{8}{15}\\





$

c)

$I_n= \dfrac{n-1}{n} * I_{n-2} \\

I_{2n+1}= \dfrac{2n+1-1}{2n+1} * I_{2n+1-2} \\
= \dfrac{2n}{2n+1} * I_{2n-1} \\
= \dfrac{(2n)*(2n-2)}{(2n+1)(2n-1)} * I_{2n-3} \\
= \dfrac{(2n)*(2n-2)*...*2}{(2n+1)(2n-1)*...*3} * I_{1} \\\\

I_1=1\\

$

3 0

3 years ago

Part C

Genrish500 [490]

<h3>Question</h3>

What are the coordinates of the vertices of the rose garden after a translation two yards east and four yards south?

Step-by-step explanation:

Answer: The coordinates of the vertices of the original rose garden are A(3, 6), B(3, 3), C(4, 3), and D(4,

8 0

2 years ago

Helppppp Plsssss Asap!! Thanks

Sonbull [250]

Answer:

Go to this link it has the answer brainly.com/question/5960663

4 0

3 years ago

Read 2 more answers

M + 5 greater than or equal to -1

sladkih [1.3K]

Answer:

m>=-6

Step-by-step explanation:

m+5>=-1

m>=-1-5

m>=-6

3 0

3 years ago