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2 years ago

15

Pls help with this it’s 3:44 AM for me, I’m in my study room trying to figure out how to solve this and I still didn’t finish My

math work cuz I didn’t know howw ( due tomorrow ) and it’s getting rlly late so can u guys help me?? ty <33! please try your best! /spamming!/

1 answer:

Degger [83]2 years ago

5 0

Answer:

1. $\frac{4}{10}$

2. He used baking soda more than vanilla flavoring. $\frac{7}{8} > \frac{1}{4}$

3. Giraffes are the herbivores. $\frac{1}{6} < \frac{3}{4}$

4. Sarai helps more in the school than Josh does. $\frac{3}{8} < \frac{8}{12}$

5. $\frac{5}{12} < \frac{6}{8}$

6. $\frac{4}{6} > \frac{2}{5}$

7. $\frac{7}{10} < \frac{4}{3}$

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Which equation has an a-value of 1, a b-value of –3, and a c-value of –5?

ANTONII [103]

Answer:

answer is first option

${x}^{2} -3x - 5 = 0$

wherein, a = 1, b = -3, c = -5

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3 years ago

Read 2 more answers

Justin lives 3 2 5 miles from his grandfather's house. Enter the mixed number as a fraction greater than one. 3 2 5 =

elena-s [515]

Answer: $\dfrac{17}{5}$ miles

Step-by-step explanation:

We can express a mixed fraction as an improper fraction.

The process is as follows:

$a\dfrac{b}{c}=\dfrac{ac+b}{c}$

The given mixed number = $3\dfrac25$ miles

This can be represented as

$3\dfrac25=\dfrac{3\times5+2}{5}\\\\=\dfrac{15+2}{5}\\\\=\dfrac{15+2}{5}\\\\=\dfrac{17}{5}>1$

Hence, the required fraction= $\dfrac{17}{5}$ miles

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2 years ago

which of these which of these is most likely to to describe the change received from a cash purchase? A.whole number B.negative

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3 years ago

No cap got a brain fart need help just gotta find x

VladimirAG [237]

Answer:

ok did you really just say brain fart I'm dying I cant stop laughing

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2 years ago

Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

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3 years ago