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3 years ago

Write an equation to represent the following statement. j is 25 more than 3

Mathematics

2 answers:

julsineya [31]3 years ago

8 0

J= 25+3
j-25=3
I'm pretty sure either would work

Harrizon [31]3 years ago

6 0

J= 25>3 I think this is how you do it let me know if it right

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I solved this problem : 33 + 45 + 78 how could i check my answer?

Vlad [161]

Step-by-step explanation:

So the answer for 33+45+78 is 156.

If you want to check that, you can use inverse operations, so subtract.

156-78=78. (Check 78)

78-45=33 (Check 45 and 33)

Well, now it's checked because we used all the factors of the equation!

4 0

3 years ago

Read 2 more answers

Find the area of a rectangle 3yds by 4/7yds

jasenka [17]

$\text{We're trying to find the area of the rectangle}\\\\\text{We know the dimensions are 3 by 4/7}\\\\\text{The formula to find the area of a rectangle is length x width}\\\\\text{Solve for area:}\\\\3\cdot\frac{4}{7}\\\\\text{Multiply the 3 to the numerator}\\\\\frac{12}{7}\\\\\text{Since we can't simply it, we'll leave it as is}\\\\\boxed{\text{Answer:} \frac{12}{7}\,\,yd^2}$

5 0

3 years ago

What is the answer? I don’t understand how to get it.

miss Akunina [59]

Answer:

Step-by-step explanation:

Sub in the (x,y) points given into the equation to see if true ....you'll find A is the answer

Example point (3,95 ) 95 = -10.375 (3) + 126.125 ??

95 = 95 Check !

then try the other point ( 11,12) to be sure ....

5 0

2 years ago

Semicircles

Studentka2010 [4]

Answer:

The area of the shaded portion of the figure is $9.1\ cm^2$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The shaded area is equal to the area of the square less the area not shaded.

There are 4 "not shaded" regions.

step 1

Find the area of square ABCD

The area of square is equal to

$A=b^2$

where

b is the length side of the square

we have

$b=4\ cm$

substitute

$A=4^2=16\ cm^2$

step 2

We can find the area of 2 "not shaded" regions by calculating the area of the square less two semi-circles (one circle):

The area of circle is equal to

$A=\pi r^{2}$

The diameter of the circle is equal to the length side of the square

$r=\frac{b}{2}=\frac{4}{2}=2\ cm$ ---> radius is half the diameter

substitute

$A=\pi (2)^{2}$

$A=4\pi\ cm^2$

Therefore, the area of 2 "not-shaded" regions is:

$A=(16-4\pi) \ cm^2$

and the area of 4 "not-shaded" regions is:

$A=2(16-4\pi)=(32-8\pi)\ cm^2$

step 3

Find the area of the shaded region

Remember that the area of the shaded region is the area of the square less 4 "not shaded" regions:

$A=16-(32-8\pi)=(8\pi-16)\ cm^2$

---> exact value

assume

$\pi =3.14$

substitute

$A=(8(3.14)-16)=9.1\ cm^2$

8 0

3 years ago

The water level varies from 12 inches at low tide to 52 inches at high tide. Low tide occurs at 9:15 a.m. and high tide occurs a

balandron [24]

52 inches - 12 inches = 40 inches
amplitude: a = 40 inches / 2 = 20

f(x)=20cos(bx)+c
the value of c is 32... since the centre of the has been moved up 32 units

the minimum amplitude = 32 - 20 = 12
the maximum amplitude = 32 + 20 = 52

f(x)=20cos(bx)+32
if the curve takes 6 1/4 hours from low to high tides (9:15 am to 3:30 pm) then it will take 12 1/2 hours to complete a full cycle.

adjust the period by converting 12 1/2 hours to an angle measure.

360°/12 = 30°
30° / 12 = 15°

12 1/2 = 360° + 15° = 375°

f(x) = 20 cos(375°) + 32
f(x) = 20 * 0.97 + 32
f(x) = 19.4 + 32
f(x) = 51.4

5 0

3 years ago