All categories

2 years ago

Please anyone help I’m stuck

Mathematics

2 answers:

worty [1.4K]2 years ago

8 0

A=10.5 maybe could be the correct answer

.

cestrela7 [59]2 years ago

5 0

Answer:

a=10.35

Step-by-step explanation:

Hey There!

All you have to do is plug in the L and W value

a= 4.5(2.3)

4.5x2.3=10.35

so a=10.35

You might be interested in

Josh will take 10 tests in 5 weeks at school. How many tests will we have taken after 7 weeks?

AveGali [126]

I don’t really know, maybe 14?

8 0

2 years ago

If the mEDG = 25°, then find the measures of the following angles.

Serga [27]

Answer:

m/BGA=25%3 EGG=GB7 EGF=CE6 BGE=BAF9

3 0

2 years ago

Find the least common multiple of 8 and 12

Elan Coil [88]

Hey there!

The LCM of 8 and 12 is 24.

Hope this helps!

8 0

2 years ago

Read 2 more answers

Find exact values for sin θ and tan θ if cos θ = -4/9 and tan θ > 0.

julsineya [31]

Answer:

Part 1) $sin(\theta)=-\frac{\sqrt{65}}{9}$

Part 2) $tan(\theta)=\frac{\sqrt{65}}{4}$

Step-by-step explanation:

we have that

The cosine of angle theta is negative and the tangent of angle theta is positive

That means that the sine of angle theta is negative

step 1

Find $sin(\theta)$

we know that

$sin^{2}(\theta) +cos^{2}(\theta)=1$

we have

$cos(\theta)=-\frac{4}{9}$

substitute

$sin^{2}(\theta) +(-\frac{4}{9})^{2}=1$

$sin^{2}(\theta) +\frac{16}{81}=1$

$sin^{2}(\theta)=1-\frac{16}{81}$

$sin^{2}(\theta)=\frac{65}{81}$

square root both sides

$sin(\theta)=\pm\frac{\sqrt{65}}{9}$

Remember that

In this problem the sine of angle theta is negative

$sin(\theta)=-\frac{\sqrt{65}}{9}$

step 2

Find $tan(\theta)$

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=-\frac{\sqrt{65}}{9}$

$cos(\theta)=-\frac{4}{9}$

substitute the given values

$tan(\theta)=-\frac{\sqrt{65}}{9}:-\frac{4}{9}=\frac{\sqrt{65}}{4}$

3 0

3 years ago

I need the answer this question hard

Genrish500 [490]

Answer:

option B is the correct answer of this question.

plz mark my answer as brainlist if you find it useful plz .hope this will be helpful to you ☺️.

6 0

2 years ago