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4 years ago

What would be the total length of the 2 1/4-inch barrettes if Mikayla places them end to end?

2 answers:

3 0

2 barrettes that measure 1/4-inch each. So we need to add them together

1/4+1/4=2/4-inch
this can be simplified to 1/2-inch

Answer=1/2-inch

d1i1m1o1n [39]4 years ago

3 0

2 barrets each 1/4 inches long equals 1/2, so 2 barrets is 1/2 inch long.

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Help! Please!

fomenos

Answer:

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Step-by-step explanation:

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3 years ago

Show with work please.

kolbaska11 [484]

Answer:

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

Step-by-step explanation:

The identity you will use is:

$\csc \left(x\right)=\frac{1}{\sin \left(x\right)}$

So,

$\csc \left(\theta-\frac{\pi }{2}\right)$

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{\sin \left(-\frac{\pi }{2}+\theta\right)}$

Now, using the difference of sin

Note: state that $\text{sin}(\alpha\pm \beta)=\text{sin}(\alpha) \text{cos}(\beta) \pm \text{cos}(\alpha) \text{sin}(\beta)$

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)}$

Solving the difference of sin:

$-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)$

$-\cos \left(\theta\right) \cdot 1+0\cdot \sin \left(\theta\right)$

$-\text{cos} \left(\theta\right)$

Then,

$\csc \left(\theta-\frac{\pi }{2}\right)=-\frac{1}{\cos \left(\theta\right)}$

Once

$\text{sec}(-\theta)=\text{sec}(\theta)$

And, $\text{sec}(\theta)=-0.73$

$-\frac{1}{\cos \left(\theta\right)}=-\text{sec}(\theta)$

$-\frac{1}{\cos \left(\theta\right)}=-(-0.73)$

$-\frac{1}{\cos \left(\theta\right)}=0.73$

Therefore,

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

3 0

3 years ago

Solve for a.

egoroff_w [7]

Answer: $a=\sqrt{\frac{30}{23} }$

Step-by-step explanation:

$23a^2=30$

Divide by 23

$a^2=\frac{30}{23}$

Extract the square root.

$a=\sqrt{\frac{30}{23} }$

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3 years ago

*Please Help!* (Easy question) 20 POINTS

mariarad [96]

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Firlakuza [10]

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