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3 years ago

I’m not sure how to do this. Plz help and show the work

Mathematics

1 answer:

kramer3 years ago

6 0

3. 65x5= 325
she traveled 325 feet in 5 minutes
4. 3(6) +5 = 23
6cube - 10 =206
5. area of the triangle is 1/2x base x height
so 1/2 x 15 x 10 = 75
7. they'll have 6 sections. 3/4 divided by 1/8 is 6

hope this helps :)

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Describe or show two ways to find the following product: 1 1/3x 2.

Alenkasestr [34]

1 and 1/3 time 2? i thinks that's what you want?

or is it 1/3x squared?

I;ll presume its 1 and 1/3 time 2

first way:

Multiple whole numbers first 1 x 2 = 2

Multiple fraction by the two 1/3 x 2 = 2/3

now add them together

2 and 2/3

second way

Make 1 and 1/3 improper = 4/3

make 2 into a a/b form = 2/1

now 4/3 x 2/1 = 8/3 now make a mixed number 2 and 2/3

8 0

3 years ago

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Find the value of a - b when a = 3 and b = -2.

lubasha [3.4K]

Answer:

Step-by-step explanation:

a = 3; b = -2

a - b = 3 - (-2) = 3 + 2 = 5

3 0

3 years ago

25% of a university's freshman class are majoring in Engineering. If 500 students in the freshman class are Engineering majors,

Pepsi [2]

Answer:

2000

Step-by-step explanation:

100/25=4

500*4=2000

4 0

4 years ago

Lease help...

m_a_m_a [10]

Bank A = 1000 x (1 + (0.04*6)) = $1,240

Bank B = 1000 x (1+0.03)^6 = $1,194.05

Bank A would be worth more

6 0

3 years ago

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Angle α lies in quadrant II , and tanα=−125 . Angle β lies in quadrant IV , and cosβ=35 .

Artist 52 [7]

Answer:

$cos(\alpha+\beta)=\frac{33}{65}$

Step-by-step explanation:

step 1

Find cos α

we know that

$tan^2(\alpha)+1=sec^2(\alpha)$

we have

$tan(\alpha)=-\frac{12}{5}$

substitute

$(-\frac{12}{5})^2+1=sec^2(\alpha)$

$sec^2(\alpha)=\frac{144}{25}+1$

$sec^2(\alpha)=\frac{169}{25}$

$sec(\alpha)=\pm\frac{13}{5}$

Remember that Angle α lies in quadrant II

sec α is negative

$sec(\alpha)=-\frac{13}{5}$

Find the value of cos α

$cos)\alpha)=\frac{1}{sec(\alpha)}$

$cos(\alpha)=-\frac{5}{13}$

step 2

Find sin α

we know that

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

$sin(\alpha)=tan(\alpha)cos(\alpha)$

we have

$tan(\alpha)=-\frac{12}{5}$

$cos(\alpha)=-\frac{5}{13}$

substitute

$sin(\alpha)=(-\frac{12}{5})(-\frac{5}{13})$

$sin(\alpha)=\frac{12}{13}$

step 3

Find sin β

we know that

$sin^2(\beta)+cos^2(\beta)=1$

we have

$cos(\beta)=\frac{3}{5}$

substitute

$sin^2(\beta)+(\frac{3}{5})^2=1$

$sin^2(\beta)=1-(\frac{3}{5})^2$

$sin^2(\beta)=1-\frac{9}{25}$

$sin^2(\beta)=\frac{16}{25}$

$sin(\beta)=\pm\frac{4}{5}$

Remember that

Angle β lies in quadrant IV

sin β is negative

$sin(\beta)=-\frac{4}{5}$

step 4

Find cos(α−β)

we know that

$cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)$

we have

$cos(\alpha)=-\frac{5}{13}$

$cos(\beta)=\frac{3}{5}$

$sin(\alpha)=\frac{12}{13}$

$sin(\beta)=-\frac{4}{5}$

substitute the given values

$cos(\alpha+\beta)=(-\frac{5}{13})(\frac{3}{5})-(\frac{12}{13})(-\frac{4}{5})$

$cos(\alpha+\beta)=(-\frac{15}{65})+(\frac{48}{65})$

$cos(\alpha+\beta)=\frac{33}{65}$

7 0

4 years ago