How many meters are in 28% of 15 km?

Help pls I have a test due tomorrow :(

Mademuasel [1]

Good luck on your test love ya also I have bad period cramps send chocolate please

5 0

3 years ago

I need help with question #2. ASAP. Anything is appreciated:)

DedPeter [7]

Y = 4/5x + 30

4/5 because the slope is 4 up, 5 right. 30 is the y-int

4 0

3 years ago

The ammunition storage room has 10 feet between the floor and the ceiling. Each box of ammunition is 10 feet tall. Each crate of

KIM [24]

Answer:

the maximum number of crates that can be stacked between the floor and ceiling

$\frac{height \hspace{0.1cm} of \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of \hspace{0.1cm} each \hspace{0.1cm} crate} = \frac{10 \hspace{0.1cm} feet}{12 \hspace{0.1cm} inches} = \frac{10 \times 12 \hspace{0.1cm} inches }{12 \hspace{0.1cm} inches} = 10 \hspace{0.1cm} crates$

where 1 foot = 12 inches. SO the answer is that a maximum of 10 crates can be stacked from floor to ceiling.

Step-by-step explanation:

i) the maximum number of crates that can be stacked between the floor and ceiling

$\frac{height \hspace{0.1cm} of \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of \hspace{0.1cm} each \hspace{0.1cm} crate} = \frac{10 \hspace{0.1cm} feet}{12 \hspace{0.1cm} inches} = \frac{10 \times 12 \hspace{0.1cm} inches }{12 \hspace{0.1cm} inches} = 10 \hspace{0.1cm} crates$

where 1 foot = 12 inches

3 0

3 years ago

If the door to the library is at (2, 4), the door to the mall is at (8, 2), and one

guapka [62]

Answer:

6 miles

Step-by-step explanation:

The distance formula is expressed as;

D = √(x2-x1)²+(y2-y1)²

D = √(2-4)²+(8-2)²

D = √(-2)²+(6)²

D = √4 + 36

AD = √40

AD = 6.32

Hence the approximate distance between the library and the mall is 6 miles

5 0

3 years ago

Can someone help me out with number 2?

Norma-Jean [14]

$\sin255^o=\sin(300^o-45^o)=\sin300^o\cos45^o-\cos300^o\sin45^o= \\\\=- \frac{ \sqrt{3} }{2}\cdot \frac{ \sqrt{2} }{2}- \frac{1}{2} \cdot \frac{ \sqrt{2} }{2}=- \frac{ \sqrt{6} }{4} - \frac{ \sqrt{2} }{4}= -\frac{ \sqrt{6}+\sqrt{2} }{4}$

5 0

4 years ago