All categories

3 years ago

Hirva ate 5/8 of a medium pizza Elizabeth ate 1/4 of the pizza how much pizza did they eat all together?

Mathematics

1 answer:

Sliva [168]3 years ago

3 0

I’m pretty sure this is 7/8!

You might be interested in

I need to know what 70% of 120 = ?

malfutka [58]

Question: <span>70% of 120 = ?

Work:

To find what </span><span>70% of 120 is. Do .70 times 120.

.70 x 120 = 84

Final Answer:

84
</span>

6 0

3 years ago

Read 2 more answers

Pls help the person with the correct answer will get a brainliest

Flauer [41]

D) Daniel’s Brick
This is because it has the most bricks close to 2.5 inches.

7 0

3 years ago

Given Triangle DEF is isosceles and angle D is the vertex angle, Find x, DE, DF, EF

ch4aika [34]

Answer:

x=2;DE,EF=9,DF=5

Step-by-step explanation:

Since DEF is isoceles, you can set the equations

x+7 and 2x+5 equal to each other

x+7=2x+5, now solve for x

-x=-2,

x=2, now plug in for each side

DE,EF=9

DF=3(2)-1=5

4 0

3 years ago

84=7(6+a) and -84= -6 (p + 6)

m_a_m_a [10]

84= 7(6+a)
84= 42 +7a
84-42= 42-42 +7a
42= 7a
42/7= 7a/7
6= a

-84= -6( p+6)
-84= -6p -36
-84+36= -6p -36+36
-48= -6p
-48/-6= -6p/-6
8 =p

7 0

3 years ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago