Ask question

Ask question

All categories

3 years ago

7

Scott is saving coins in a jar. Last week, the jar was 1/4 full. This week, he added more coins, and the jar is now 1/2 full. Ho

w much of the jar did Scott fill this week?

2 answers:

Margaret [11]3 years ago

7 0

Answer:

He filled his jar 1/4 more this week.

Step-by-step explanation:

The explanation is very simple, you know that he added 1/4 coins last week and this week he added more coins that filled the jar half way:

1/2 - 1/4 = 1/4

or you can also use:

1/4 + 1/4 = 1/2

Leona [35]3 years ago

4 0

Answer:

Scott filled (1/2 - 1/4) of the jar, or 1/4 of the jar this week.

Let me know if this helps!

You might be interested in

The difference of 10 and a number w is no more than 8

Tpy6a [65]

Answer:10-w<u>< </u> 8

Step-by-step explanation:

8 0

3 years ago

How do I determine z ∈ C:

saw5 [17]

Simplify the coefficient of z on the left side. We do this by rationalizing the denominators and multiplying them by their complex conjugates:

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3-2i}{1+i}\cdot\dfrac{1-i}{1-i} - \dfrac{5+3i}{1+2i}\cdot\dfrac{1-2i}{1-2i}$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{(3-2i)(1-i)}{1-i^2} - \dfrac{(5+3i)(1-2i)}{1-(2i)^2}$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3 - 2i - 3i + 2i^2}{1-(-1)} - \dfrac{5 + 3i - 10i - 6i^2}{1-4(-1)}$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3 - 5i + 2(-1)}2 - \dfrac{5 - 7i - 6(-1)}5$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{1 - 5i}2 - \dfrac{11 - 7i}5$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{1 - 5i}2\cdot\dfrac55 - \dfrac{11 - 7i}5\cdot\dfrac22$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{5 - 25i - 22 + 14i}{10}$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = -\dfrac{17 + 11i}{10}$

So, the equation is simplified to

$-\dfrac{17+11i}{10} z = \dfrac12 - \dfrac{2i}5$

Let's combine the fractions on the right side:

$\dfrac12 - \dfrac{2i}5 = \dfrac12\cdot\dfrac55 - \dfrac{2i}5\cdot\dfrac22$

$\dfrac12 - \dfrac{2i}5 = \dfrac{5-4i}{10}$

Then

$-\dfrac{17+11i}{10} z = \dfrac{5-4i}{10}$

reduces to

$-(17+11i) z = 5-4i$

Multiply both sides by -1/(17 + 11i) :

$\dfrac{-(17+11i)}{-(17+11i)} z = \dfrac{5-4i}{-(17+11i)}$

$z = -\dfrac{5-4i}{17+11i}$

Finally, simplify the right side:

$-\dfrac{5-4i}{17+11i} = -\dfrac{5-4i}{17+11i} \cdot \dfrac{17-11i}{17-11i}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{(5-4i)(17-11i)}{17^2-(11i)^2}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{85 - 68i - 55i + 44i^2}{289-121(-1)}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{85 - 68i - 55i + 44(-1)}{410}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{41 - 123i}{410}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{41 - 41\cdot3i}{410}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{1 - 3i}{10}$

So, the solution to the equation is

$z = -\dfrac{1-3i}{10} = \boxed{-\dfrac1{10} + \dfrac3{10}i}$

4 0

3 years ago

Write the equation you would need to solve to find the horizontal distance each beam is from the origin

anzhelika [568]

Answer:

the red line: $y\ =\ \sqrt{900-x^{2}}$

the black line: $y=\frac{1}{27.5}x^{2}+15$

Below the graph.

6 0

3 years ago

Jessie designed a sculpture that is shaped like a circle. The circumference of the sculpture is 10π meters. Which measurement is

BlackZzzverrR [31]

Answer:

25πm²

Step-by-step explanation:

Circumference of the circular sculpture = πd

d is the diameter

Given

Circumference = 10π metres

Substitute into the formula

10π = πd

10 = d

d = 10m

Area of the sculpture = πd²/4

area of the sculpture = π(10)²/4

area of the sculpture = 100π/4

area of the sculpture = 25π m²

Hence the measurement closest to the area of the sculpture in square meters is 25πm²

6 0

3 years ago

1. Find the vertices and locate the foci for the hyperbola whose equation is given.

Irina18 [472]

$\bf \cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1
\qquad center\ ({{ h}},{{ k}})\qquad
 vertices\ ({{ h}}\pm a, {{ k}})\\\\
-----------------------------\\\\
\textit{now let's take a look at yours}
\\\\\\
49x2 - 16y2 = 784\implies \cfrac{49x^2}{784}-\cfrac{16y^2}{784}=1
\\\\\\
\cfrac{x^2}{16}-\cfrac{y^2}{49}=1\implies \cfrac{(x-0)^2}{4^2}-\cfrac{(y-0)^2}{7^2}=1
\\\\\\
recall\implies center\ ({{ h}},{{ k}})\qquad vertices\ ({{ h}}\pm a, {{ k}})
\\\\\\
$

$\bf \textit{now, for the foci, the foci are "c" distance from the center point}\\\\\
whereas\qquad c=\sqrt{a^2+b^2}\qquad \textit{ that is }\qquad h\pm \sqrt{a^2+b^2}$

notice your "a" and "b" components, to get the distance "c" from the center to either foci and the vertices, of course, are h + a, k and h - a, k

5 0

3 years ago