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2 years ago

Please answer this correctly

Mathematics

2 answers:

Artemon [7]2 years ago

3 0

Answer:

1 flowers

Step-by-step explanation:

The graph shows that only one flower received more than $4\frac{1}{2}$ cups of water and that is the plant that received 5 cups of water.

grandymaker [24]2 years ago

3 0

1 flower

Step-by-step explanation:

the question asks for flowers above the 4 1/2 mark and only 1 flower is there.

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The correct option is the third option. -2 1/4 / -2/3 = 3 3/8

-9/4 / -2/3 = 3 3/8

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Complete the square to the problem X^2+6x-16=0

marishachu [46]

Answer:

Step-by-step explanation:

$(a+b)^2=a^2+2ab+b^2\qquad(*)\\\\-------------------------------\\\\x^2+6x-16=0\qquad\text{add 16 to both sides}\\\\x^2+6x=16\\\\x^2+2(x)(3)=16\qquad\text{add}\ 3^2\ \text{to both sides}\\\\\underbrace{x^2+2(x)(3)+3^2}_{(*)}=16+3^2\\\\(x+3)^2=16+9\\\\(x+3)^2=25\Rightarrow x+3=\pm\sqrt{25}\\\\x+3=-5\ \vee\ x+3=5\qquad\text{subtract 3 from both sides}\\\\\boxed{x=-8\ \vee\ x=2}$

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What is the volume of the right rectangular prism? Enter your answer in the box. cm³ A rectangular prism with a length of 6.1 ce

Vanyuwa [196]

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PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

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

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

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Lostsunrise [7]

Answer:

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1/8+2/8= 3/8

0/8<3/8<4/8

Step-by-step explanation:

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