All categories

3 years ago

Kevin ran for less than 5 miles writ an inequality for each sentence.

Mathematics

2 answers:

ryzh [129]3 years ago

8 0

$k \leq 5$
k=kevin's running

Vesnalui [34]3 years ago

5 0

K<5

Because it says less than which means it that the sign we are using is <

You might be interested in

Help me plz !!!!!!!!!!!!!!!!!

trapecia [35]

Answer:

the answer should be 54.6 grams

7 0

3 years ago

What is the value of the expression below when w =9 and x=2<br><br> w – 4x

Vera_Pavlovna [14]

Answer:

Step-by-step explanation:

9 - 4 x 2

4 x 2 = 8

9 - 8 = 1

8 0

3 years ago

Read 2 more answers

I WILL GIVE CORRECT ANSWER BRAINLIEST!

vesna_86 [32]

Step-by-step explanation:

In Kate's equation, 20 is the total number of tickets, 4 is number of tickets per ride, r is number of rides already ridden, and 4r is the number of tickets used

in Brian's equation, 5 is total number of rides Katie can ride and 5 - r is the number of rides left to ride

3 0

3 years ago

Ann increased the quantities of all the ingredients in a recipe by 60% She used 80 grams (g) of cheese How much cheese did the r

Delvig [45]

Answer:g = 50 g

Step-by-step explanation:

She used 60% more than required ⇒ she used (1+0.60)=1.60 times what was required.

x = grams required by the recipe

80 g = 1.60x

x = (80/1.60) g = 50 g

3 0

3 years ago

Find the area ratio of a regular octahedron and a tetrahedron regular, knowing that the diagonal of the octahedron is equal to h

Anton [14]

Answer:

$\frac{4}{3}$

Step-by-step explanation:

The area of a regular octahedron is given by:

area = $2\sqrt{3}\ *edge^2$ . Let a is the length of the edge (diagonal).

area = $2\sqrt{3}\ *a^2$

Given that the diagonal of the octahedron is equal to height (h) of the tetrahedron i.e.

a = h, where h is the height of the tetrahedron and a is the diagonal of the octahedron. Let the edge of the tetrrahedron be e. To find the edge of the tetrahedron, we use:

$h=\sqrt{\frac{2}{3} } e\\but\ h=a\\a=\sqrt{\frac{2}{3} } e\\e=\sqrt{\frac{3}{2} }a$

The area of a tetrahedron is given by:

area = $\sqrt{3}\ *edge^2$ = $\sqrt{3} *(\sqrt{\frac{3}{2} }a)^2=\frac{3}{2}\sqrt{3} *a^2$

The ratio of area of regular octahedron to area tetrahedron regular is given as:

Ratio = $\frac{2\sqrt{3}\ *a^2}{\frac{3}{2} \sqrt{3}*a^2} =\frac{4}{3}$

7 0

3 years ago