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3 years ago

Add 8/9+4/9 and Simplify the answer and write as a mixed number.

Mathematics

2 answers:

Galina-37 [17]3 years ago

7 0

Answer:

1 1/3

Step-by-step explanation:

fenix001 [56]3 years ago

6 0

Answer:

12/9= 1 and 1/3

Step-by-step explanation:

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4-1/x divided by 16-1/xsquared

Vladimir [108]

4 - 1/x (16)-1/x 2
4x-18/x

3 0

3 years ago

A rectangle has a length of 4 mm has an area is 44 mm² width of the rectangle is

nordsb [41]

The answer is : w=1.1×10^7mm

5 0

2 years ago

Read 2 more answers

Please answer the problem below.<br><br> ~Thanks = )

RoseWind [281]

Hi GlaxyCraft4991,

Solution:

−12x − 0.4 > 0.2 (36.5x + 80) − 55

−12x − 0.4 > 7.3x − 39

−12x − 0.4 − 7.3x > 7.3x − 39 − 7.3x

−19.3x − 0.4 > −39

−19.3x − 0.4 + 0.4 > −39 + 0.4

−19.3x > −38.6

−19.3x/19.3 > −38.6/19.3

x < 2

Correct Answer:

x < 2

3 0

3 years ago

Find the volume and area for the objects shown and answer Question

klio [65]

Step-by-step explanation:

You must write formulas regarding the volume and surface area of the given solids.

$\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2$

$\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\$

$\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}$

$\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}$

$SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}$

6 0

3 years ago

What is the current density of the solution a/m2

Jlenok [28]

Current density of solution a/m2 is given below

Step-by-step explanation:

Current density is the measure of current generated per unit surface area of electrode.

Eg: In microbial fuel cell/fuel cells, if you are using electrode with size of 5cm x 2cm and current produced is 200mA, then current density will be,

CD= Current/(projected area of electrode)

area of rectangle= LxB= 5cm x 2cm =10cm2

CD= 200mA/10cm2

CD= 20mA/cm2

Your Current will be 200mA.

Current density will be 20mA/cm2.

8 0

3 years ago