For art class, Mia needs to cut out triangles that take up exactly 9 cm². Which is a possible measurement?

A parallelogram has a base of 10 centimeters and a height of 4 centimeters. What is the area of the parallelogram?

Illusion [34]

Step-by-step explanation:

Area of a parallelogram=base×height

=10×4

=40cm^2

5 0

2 years ago

SOMEEEEONEEE pls help pls and thx :)

alekssr [168]

1) The sculptor created a marble basin with an approximate volume of 94,509.6 cubic centimeters - Option D, 2) The approximate area of the heart-shaped cake is 283 square inches - Option B, 3) The length of the wooden frame around the window is 94.3 inches - Option B.

In this question we should make use of geometric formulas for volumes and areas and key information from statement in order to find the right choices.

1) In this case, the volume occupied by the marble water basin ( $V$ ), in cubic centimeters, by subtracting the volume of the hemisphere from the volume of the cylinder:

$V = \pi\cdot R^{2}\cdot h - \frac{2\pi}{3} \cdot r^{3}$ (1)

Where:

$R$ - Radius of the cylinder, in centimeters.
$h$ - Height of the cylinder, in centimeters.
$r$ - Radius of the cylinder, in centimeters.

If we know that $R = 30\,cm$ , $h = 45\,cm$ and $r = 25\,cm$ , then the volume of the marble is:

$V = \pi \cdot (30\,cm)^{2}\cdot (45\,cm) - \frac{2\pi}{3}\cdot (25\,cm)^{3}$

$V \approx 94509.579\,cm^{3}$

The right choice is D.

2) To determine the approximate surface area of the cake covered in red frosting ( $A_{s}$ ), in square inches, we need to find the sum of the surface area of the entire circle and the surface area of the square:

$A_{s} = l^{2} + 2\cdot l\cdot h + \frac{\pi}{4}\cdot D^2 + \pi \cdot D\cdot h$ (2)

Where:

$l$ - Side length of the square, in inches.
$h$ - Height of the cakes, in inches.
$D$ - Diameter of the cakes, in inches.

If we know that $l = 9\,in$ , $h = 3\,in$ and $D = 9\,in$ , then the approximate area covered in red frosting:

$A_{s} = (9\,in)^{2} + 2\cdot (9\,in)\cdot (3\,in) + \frac{\pi}{4}\cdot (9\,in)^{2} + \pi \cdot (9\,in)\cdot (3\,in)$

$A_{s} \approx 283.440\,in^{2}$

The right choice is B.

3) The frame around the window is found by means of the following perimeter formula ( $s$ ), in inches:

$s = \pi\cdot r + 2\cdot h + 2\cdot r$ (3)

Where:

$r$ - Radius, in inches.
$h$ - Height of the rectangle, in inches.

If we know that $r = 9\,in$ and $h = 24\,in$ , then the length of the frame around the window is:

$s = \pi\cdot (9\,in) + 2\cdot (24\,in) + 2\cdot (9\,in)$

$s \approx 94.274\,in$

The right choice is B.

We kindly invite to see question on volumes: brainly.com/question/1578538

7 0

2 years ago

A yoga studio offers memberships that cost $60 per month for unlimited classes. The studio also accepts walk-ins, charging $6 pe

Svetach [21]

Answer:

10 classes

Step-by-step explanation:

Ok, so to solve this we have to make two different equations. we can say 60x = y, or $60 per month, and 6a = b, or $6 per walk in. Since we want the options to end up costing the same, we can say that y = b. With substitution, we get that 60x, (60 per month) = 6a (6 per drop in). The question says we have 1 month, so x = 1. This means we have 60 = 6a, and a = 10 (10 classes).

7 0

2 years ago

Read 2 more answers

A mathematics teacher wanted to see the correlation between test scores and homework. The homework grade (x) and test grade (y)

Fiesta28 [93]

Answer:

Il existe de nombreux exemples de portfolios d’artistes en ligne. Trouvez-en un et partagez un lien vers celui-ci. En vous basant sur le portfolio, écrivez quelques phrases évaluant l'artiste. S'il y a les réponses d'autres élèves sur le forum de discussion, examinez-les et voyez si vous êtes d'accord ou pas d'accord avec leur opinion sur le travail des artistes.

6 0

1 year ago

An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were

vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

Step:-(1)

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

Step :-2

Null hypothesis: H₀: there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Alternative hypothesis :H₁: there is significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

$Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} } } }$

Given n₁=n₂=60.

$Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60} } }$

On calculation, we get

Z = $\frac{4}{\sqrt{0.6833} }$

z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

3 0

2 years ago