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adell [148]
1 year ago
12

Round the number 76.4491 to 1 decimal place

Mathematics
1 answer:
solniwko [45]1 year ago
4 0

Answer:

Step-by-step explanation:

76.4

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I am choosing shapes out of a bag. The ratio of blue shapes to red shapes are 2:1. For the blue shapes, the ratio of spheres to
Allisa [31]

Answer:

18

Step-by-step explanation:

as there are 12 red spheres there are 12*4 red cubes (48) add them ,60.Then 60/2 to find blue shapes ,30, then add the 2 and 3 to make 5 after that do 30/5 = 6 then multiply by 3

8 0
1 year ago
Cate pahare de 10 cl se pot umple ce sucul dintr-o cutie de 100 cl?
Bumek [7]
English plzz ? I don't understand Spanish
3 0
1 year ago
A ball is kicked 4 feet above the ground with an initial vertical velocity of 51 feet per second. The function h(t)=−16t2+51t+4
Katyanochek1 [597]

Check the first picture below.

now, the exercise states to get it from the graph, so we have to graph it first.

Check the second picture below.

recall, the height in feet is on the y-axis, and the time in seconds is on the x-axis.

and rounded up to the nearest tenth will be 0.6.

7 0
1 year ago
Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
1 year ago
Read 2 more answers
A manageress earned £24,500 last year. This year she earns £25,235.
Korolek [52]

Answer:

Step-by-step explanation:

245000 last year

This year 25235

(y2 - y1) / y1)*100 = your percentage change

(where y1=start value and y2=end value)  

(( £25.235 - £24.500) / £24.500) * 100 = 0 %

There ain't no percentage change as there needs to be a bigger difference between the two numbers plus u should use the formula

7 0
1 year ago
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