All categories

4 years ago

A liter is equal to 100 centiliters.Is a centiliter a greater measure than a milliliter. Explain

Mathematics

2 answers:

neonofarm [45]4 years ago

8 0

True with that answer

borishaifa [10]4 years ago

6 0

Yes. It takes more milliliters than centiliters to fill a liter.

You might be interested in

Significa lo mismo el .1 en 20.1 km que en 5.1 <br> por que?

REY [17]

Answer:

Si, en tanto los kilómetros siguen un patrón decimal, al igual que los números enteros.

Step-by-step explanation:

La pregunta refiere a si el .1 implica un mismo valor en 20.1 km que en 5.1, es decir, en un numero sin unidad de medición.

Y la respuesta es positiva, en tanto los kilómetros, al igual que los números enteros, se rigen por un patrón decimal, es decir, donde los números pueden descomponerse en unidades divisibles por diez. Así, el .1 implica que, tanto en el kilómetro como en el numero entero, se trata del 10% del valor de la unidad en cuestión.

5 0

4 years ago

Simplify the expression x^2+3x-28/x^2-7x+12

EleoNora [17]

X^2 + 3x - 28 factors to (X+7)(X-4)
X^2 - 7X + 12 factors to (X-4)(X-3)
Now we see that the equation is [(X+7)(X-4)]/[(X-4)(X-3)]
By dividing the top and bottom by (X-4), we see that the equation equals (X+7)/(X-3).

6 0

3 years ago

Read 2 more answers

When 50 pounds of a compound are produced, the manufacturing cost is described by x

Paraphin [41]

Answer:

Step-by-step explanation:

i don't know

4 0

3 years ago

sierra worked for 38 hours this week and earned $427.50. write and solve an equation determine how much she makes per hour

Kisachek [45]

Answer:

$11.25 an hour

Step-by-step explanation:

427.50=38x

38x=427.5

÷by 38

x=11.25

3 0

3 years ago

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term $2x^2$ , which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression:

$x_v=\frac{-b}{2a}$

Since in our case $a=2$ and $b=-3$ , we get that the x-position of the vertex is:

$x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

$y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}$

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

$f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6$

See the graph produced in the attached image.

4 0

3 years ago