All categories

3 years ago

Solve using method of your choice. PLease help 28 points

1 answer:

5 0

$7x+y=-20\\
x-3y=-50\\\\
21x+3y=-60\\
\underline{x-3y=-50}\\
22x=-110\\
x=-5\\\\
-5-3y=-50\\
3y=45\\
y=15
$

-----------------------------------------------------

$(5x-8y)^2=-25$
No solution in real numbers.

You might be interested in

ASSIGNMENT: Show that the correct solution to the following system of equations is (-2, 0). Upload a photo of your work or neatl

ki77a [65]

Answer:

Step-by-step explanation:

5 0

3 years ago

A 4.5 kg cat requires medication at a dose rate of 5 mg/kg once daily for the next 7 days. The tablets are available in 10 mg st

Marta_Voda [28]

Answer:

Approximately 16 tablets of medicine

Step-by-step explanation:

Based on the information above , we can use a Rule of Three to first figure out how many mg the cat would need to ingest per day.

$\frac{5mg}{1kg} = \frac{x}{4.5kg}$

$\frac{5mg*4.5kg}{1kg} = 22.5mg$

Now we see that the cat needs to take a daily dose of 22.5mg. Since the Tablets come at a strength of 10 mg per tablet then,

$22.5 / 10 = 2.25$

The cat would need to take 2.25 tablets per day. Since she needs to take the dose for the next 7 days then,

$2.25 * 7 = 15.75$

She would need to take approximately 16 tablets of medicine during the week .

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

5 0

3 years ago

What is the solution to this equation -3x=52

lisov135 [29]

Answer:

-17.3

Step-by-step explanation:

To solve for x, we can divide both sides by -3:

x = -52 / 3 which = -17.3

8 0

3 years ago

Read 2 more answers

What is the value of b?

labwork [276]

The value of b is 61°.

you solve by doing this:
add up the two given angles, 58° and 61° to get 119°. then you subtract from 180° to get 61.

4 0

3 years ago

Read 2 more answers

Helppp me plsssssssss

Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

$\sf \:\:\:\:\:\:\:\:\:\:x_{k}=35$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k}=50$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42$
$\sf \:\:\:\:\:\:\:\:\:\:h=5$

${\bf \:\: {By\:using\:the\: formula}} \\ \\$

$\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\$

$\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\$

${\large{\frak{\pmb{\underline{Additional\: information }}}}}$

MODE

Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

In a frequency distribution the class having maximum frequency is called the modal class.

${\bf{\underline{Formula\:for\: calculating\:mode:}}} \\$

${\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\$

Where,

$\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.$

$\small \blue{ \bigstar}$ $\sf \: f_{k}=frequency\:of\:the\:modal\:class$

$\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class$

$\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class$

$\small \purple{ \bigstar}$ $\sf \: h= width \:of\:the\:class\:interval$

7 0

3 years ago