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3 years ago

Can u help me plz ?

Mathematics

1 answer:

natita [175]3 years ago

8 0

Answer:

82 grams

Step-by-step explanation:

each rubber band weighs 3 grams because when 2 bands were added to 88 grams it went up 6 grams and six divided by 2 equals 3.

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Given log4(3) =0.7925 and log4(5) =1.1610, use log properties to find log4(300).

Ede4ka [16]

Answer:

4.1145

Step-by-step explanation:

You can use the following formula to solve this kind of problems,

$log_{a}(b) + log_{a}(c) = log_{a}(bc)$

So, let's start of with log4_(300). Since the question have already gave log4_(3) =0.7925 and log4_(5) =1.1610. It means that you have to split the 300 into the simplest form where almost all of the numbers are 3 and 5.

$log_{4}(300) = log_{4}(4 \times 5 \times 3 \times 5) \\ = log_{4}(4) + log_{4}(5) + log_{4}(3) + log_{4}(5) \\ = 1 + 1.1610 + 0.7925 + 1.1610 \\ = 4.1145$

5 0

3 years ago

PLS HELP BEING TIMED!!!!

artcher [175]

Answer:

the second choice cause there's too many outliers the line can't be linear

7 0

3 years ago

Please help will mark brainliest

Butoxors [25]

4, 16
-1, 1
3, 9
12, 144
16, 256

The rule is y=x^2. If you square all the numbers on the x side, you will get the numbers on the y side. 4 times 4 = 16. -1 times -1 =1. 3 times 3 =9, and so on.

5 0

3 years ago

A circle is centered at the point (-7, -1) and passes through the point (8, 7). The radius of the circle is units. The point (-1

padilas [110]

hello :
an equation of the circle Center at the A(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : a = -7 and b = -1 (Center at: A(-7,-1) )
r = AP.... P(8,7)
r² = (AP)²
r² = (8+7)² +(7+1)² =225+64=289 ...... so : r = 17
an equation of the circle that satisfies the stated conditions.
Center at A(-7,-1), passing through P(8, 7) is :

(x+7)² +(y+1)² = 289

The point (-15,y ) lies on this circle : (-15+7)² +(y+1)² = 289....(subsct : x= -15)
(y+1)² = 225
(y+1)² = 15²
y+1 = 15 or y+1 = -15
y = 14 or y = -16
you have two points : (-15,14) , (-15, -16)

8 0

3 years ago

You work part-time at the “Pack-it and Mail-it” that sells containers used for shipping items. Your photography teacher asks you

Lena [83]

Answer:

(a)18.85 Cubic Inches

(b)The box with dimensions of 4 in. x 3in. x 2 in.

Step-by-step explanation:

Part A

Volume of the 12 Containers

Height =2 Inches

Diameter=1 Inch

Radius=Diameter/2=1/2=0.5 Inch

Volume of a cylinder $=\pi r^2 h$

Volume of the 12 Containers

$=12 X \pi r^2 h\\=12 X\pi X0.5^2X2\\=6\pi $ cubic inches\\=18.85 \text{cubic inches (to two decimal places)}$

Part B

To determine the container which should be used, we first determine the volume of the available boxes.

Volume of the boxes

Volume of box with dimension 3 in. x 2 in. x 5 in.=3X2X5=30 Cubic Inches

Volume of box with dimension 4 in. x 3in. x 2 in.=4X3X2=24 Cubic Inches

Volume of box with dimension 5 in. x 6 in. x 5 in.=5X6X5=150 Cubic Inches

Volume of box with dimension 3 in. x 2 in. x 3 in. =3X2X3=18 Cubic Inches

The box that should be used with the least amount of wasted space is he box with volume 24 Cubic inches. i.e. box with dimensions 4 in. x 3in. x 2 in.

This is because the volume of the box(18 cubic inches) with dimension 3 in. x 2 in. x 3 in. is less than what is required.

4 0

3 years ago