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3 years ago

13

Please help! -10x

dle" class="latex-formula">-100

1 answer:

Tresset [83]3 years ago

8 0

Get the unknown by itself: Divide both sides by -10: X≤10. Since you divided by a negative number the less than sign became a greater than sign. I hope that makes sense :)

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Answer:

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3 years ago

The diagonals of a rhombus measures 6cm and 8cm, what is the perimeter of this rhombus?

Kamila [148]

<h2><u>Diagram</u><u>:</u><u>-</u></h2>

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A}\put(6,0.5){\sf B}\put(1.4,4.3){\sf D}\put(6.6,4.3){\sf C}\end{picture}$

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

The diagonals of the Rhombus= d1 & d2=6cm and 8cm

Let the Side=a

As we know that in a Rhombus

${\boxed{\sf side=\sqrt {\left ({\dfrac{d1}{2}}\right)^2+\left ({\dfrac {d2}{2}}\right)^2}}}$

Substitute the values

${:}\longmapsto$ $\sf a=\sqrt {\left ({\dfrac {6}{2}}\right)^2+\left ({\dfrac {8}{2}}\right)^2 }$

${:}\longmapsto$ $\sf a=\sqrt {(3)^2+(4)^2}$

${:}\longmapsto$ $\sf a=\sqrt {9+16}$

${:}\longmapsto$ $\sf a=\sqrt{25}$

${:}\longmapsto$ $\sf a=5cm$

<h2>_____________</h2><h3>Again </h3>

we know that in a Rhombus

$\boxed{\sf Perimeter=4a}$

Substitute the values

${:}\longmapsto$ $\sf Perimeter=4×5$

${:}\longmapsto$ $\sf Perimeter=20cm$

$\therefore$ Perimeter of the rectangle is 20cm.

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2 years ago