Help I will be marking brainliest!!!

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Over [174]

Volume of cuboid =l×b×h

area of base of the cuboid =l×b

hence

volume of cuboid is = area of base × height

140=88×h

h=140÷88

h=5

4 0

3 years ago

Approximate solution to this equation?

diamong [38]

I don’t know the answer i just want points

8 0

3 years ago

Read 2 more answers

Which function has a range of all real numbers greater than or equal to -4?

max2010maxim [7]

Answer:

C. $f(x)=(x+1)^2-4$

Step-by-step explanation:

A. the graph is a straight line with slope 1. It goes up/down infinitely far so the range is all real numbers. Not A!

B. The graph is a straight line with slope -4, so, like the function in A, the range is all real numbers. Not B!

D. The graph is an absolute value function y = |x|, reflected over the x-axis, shifted right 3 units, then shifted down 4 units. So the graph starts witha "vee" shape opening up, becomes a vee opening <u>down</u>, then ultimately gets shifted down 4 units. The range is all real numbers less than or equal to -4.

See the attached graphs. image2 is the function in answer choice D.

7 0

3 years ago

A bag contains counters that are red, black , or green . 1/3 of the counters are red 1/6 of the counters are black There are 15

Nadya [2.5K]

Answer:

There are 5 black counters in the bag.

Step-by-step explanation:

15 green counters in the bag

The proportion of green counters is given by:

$p = 1 - (\frac{1}{3} + \frac{1}{6}) = 1 - (\frac{2}{6}+\frac{1}{6}) = 1 - \frac{3}{6} = 1 - \frac{1}{2} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2}$

So, we have that, the total is x. So

$\frac{x}{2} = 15$

$x = 30$

There are 30 total counters.

How many black counters are in the bag ?

A sixth of the counters are black. So

$\frac{1}{6} \times 30 = \frac{30}{6} = 5$

There are 5 black counters in the bag.

8 0

3 years ago

Which of the following is a characteristic of a regular tessellation?

Svetach [21]

<h3>Answer: Choice A</h3>

Explanation:

For regular tessellations, the types of polygons that we can use are

squares or rectangles
equilateral triangles
regular hexagons

We can only pick one of those shapes.

Shapes like regular octagons or pentagons will not work on their own because there is gap or overlap if we tried to glue them together. Think of tiles on the floor. There cannot be any gap or overlap when forming a tessellation.

If you wanted to use octagons to tessellate the plane, then you'd need squares to fill in the gaps. At this point, it's not considered a regular tessellation.

5 0

3 years ago